Static Metallic Manual - Mts004

Technical Manual MTS 004 Iss. C External distribution authorised:

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Static stress manual, metallic materials

1 4

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Volume 1

5

Structural Design Manuals

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Contents

Document Manager

Methods for calculating static failure loads and stresses for aircraft metallic structural details.

All programmes, static justification of metallic structures.

Not applicable.

V1 - 1 V1 - 2 V1 - 3 V1 - 4 V1 - 5 V1 - 6 V1 - 7 V1 - 8 V1 - 9

Dept code: BTE/CC/CM

Stiffened panels Buclking of plates and thin shells Stiffeners Thin web beams Stable web beams Bolted or rivetted junctions Lugs Hole reinforcements Stabilisers

Validation

Name: J. HUET

Name: JF. IMBERT Function : Deputy Department Group Leader Dept code: BTE/CC/A Date: 11/99 Signature

This document is the property of AEROSPATIALE MATRA AIRBUS; no part of it shall be reproduced or transmitted without authorization of AEROSPATIALE MATRA AIRBUS and its contents shall not be disclosed. © AEROSPATIALE MATRA AIRBUS - 1999

4page 1

Title - Annex

Reference documents

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Abbreviations

C BE 019: Drawing up of the Structural Justification Dossier

See bibliography at the beginning of each chapter.

See Lexique Aerospatiale Airbus/ATR See "General" paragraph of each chapter

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List of words the definitions of which are integrated into the Lexique Aerospatiale Airbus/ATR:

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02/98

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Changes as per table page V1-7.i.

V1 - 4

New chapter.

V1 - 1


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11/99

Justification of the changes made New document.

Created paragraph V1-1-8. V1 - 5

© AEROSPATIALE MATRA AIRBUS - 1999

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MTS 004 Iss. C

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MTS 004 Iss. C

page IG1

MCS V1-1 • STIFFENED PANELS CONTENTS issue V1-1 V1-1 V1-1 V1-1 V1-1 V1-1 V1-1 V1-1 V1-1

STIFFENED PANELS 1 General 2 Preparing basic data 3 Pocket folding stresses 4 Calculating stiffened panels single compression 5 Calculating flat stiffened panels single nominal shear 6 Calculating curved stiffened panels single nominal shear 7 Calculating stiffened panels compression and shear 8 Calculating stiffened panels compression and bending

2 1 1 2 2 2 2 2 1

date

change

8/1999 Creation V1-1 8 Creation 1/1998 Creation 1/1998 Creation 1/1998 Creation 1/1998 Creation 1/1998 Creation 1/1998 Creation 1/1998 Creation 8/1999

1 GENERAL 1.1- Definitions (V1-11/1)* 1.2- Behaviour, failure modes (V1-11/3) 1.3- Calculation methodology (V1-11/5) 1.4- Scope (V1-11/6)

2 PREPARING BASIC DATA 2.1- Material characteristics (V1-12/1) 2.2- Super-stiffener sections (V1-12/2) 2.3- Example (V1-12/5)

3 POCKET FOLDING STRESSES 3.1- Hypotheses (V1-13/1) 3.2- Boundary conditions (V1-13/2) 3.3- Design 3.4- Example (V1-13/3)

4 CALCULATING STIFFENED PANELS SINGLE COMPRESSION 4.1- Principles (V1-14/2) 4.2- Widths and load-carrying section (V1-14/3) 4.3- Effective buckling length, clamping factor (V1-14/4) 4.4- Allowable stresses at UL (V1-14/6) 4.4.1 Local buckling 4.4.2 Crippling, lateral buckling, allowable stress at zero slenderness ratio in stiffener (V1-14/7) 4.4.3 Allowable stress at zero slenderness ratio in super-stiffener 4.5- Calculating margin at UL using Engesser formula (V1-14/8) 4.5.1 Principle 4.5.2 Modified Engesser formula 4.5.3 Limits, accuracy 4.5.4 Methodology 4.5.5 Example (V1-14/10) *: Page number between brackets

Issue 2  AEROSPATIALE 1999

Contents

page •i

MCS V1-1 • STIFFENED PANELS 4.6- Calculating margin at UL using Johnson formula (V1-14/15) 4.6.1 Principle 4.6.2 Limits, accuracy 4.6.3 Methodology (V1-14/16) 4.6.4 Example (V1-14/18) 4.7- Simplified approach to the Johnson formula (V1-14/22) 4.7.1 Method 4.7.2 Example (V1-14/23)

5 CALCULATING FLAT STIFFENED PANELS SINGLE NOMINAL SHEAR 5.1- Diagonal tension theories (V1-15/2) 5.2- Limits (V1-15/6) 5.3- Calculation principle for stiffened panels incomplete diagonal tension (V1-15/7) 5.4- Loading ratio, diagonal tension factor (V1-15/9) 5.5- Widths and load-carrying sections 5.6- Stresses in super-stiffener (V1-15/11) 5.6.1 General method 5.6.2 Simplified approaches (V1-15/12) 5.7- Stresses in skin (V1-15/14) 5.7.1 In pocket 5.7.2 At stiffener 5.8- Stresses in stiffener (V1-15/15) 5.9- Allowable stresses at UL 5.9.1 Local buckling 5.9.2 Forced crippling 5.9.3 Buckling of super-stiffener (V1-15/16) 5.9.4 Skin failure 5.10- Skin stiffness 5.11- General instability, design (V1-15/17) 5.12- Example (V1-15/18)

6 CALCULATING CURVED STIFFENED PANELS SINGLE NOMINAL SHEAR 6.1- Diagonal tension theories (V1-16/2) 6.2- Limits (V1-16/5) 6.3- Calculation principle for stiffened panels incomplete diagonal tension (V1-16/6) 6.4- Loading ratio, diagonal tension factor (V1-16/7) 6.5- Widths and load-carrying sections (V1-16/8) 6.5.1 Stiffener orientation 6.5.2 Frame orientation (V1-16/10) 6.6- Stresses in super-members 6.7- Stresses in skin (V1-16/12) 6.7.1 In pocket 6.7.2 At stiffener 6.8- Stresses in stiffener (V1-16/13)

page •ii  AEROSPATIALE 1999

Contents

Issue 2

MCS V1-1 • STIFFENED PANELS 6.9- Stresses in frames (V1-16/14) 6.10- Allowable stresses at UL 6.10.1 Local buckling 6.10.2 Forced crippling 6.10.3 Buckling of super-stiffener 6.10.4 Skin failure 6.11- Skin stiffness (V1-16/15) 6.12- General instability, local buckling of frames, design 6.13- Example (V1-16/17)

7 CALCULATING STIFFENED PANELS COMPRESSION AND SHEAR 7.1- Calculation principles (V1-17/2) 7.2- Limits (V1-17/4) 7.3- Loading ratio, diagonal tension factor 7.3.1 Loading ratio 7.3.2 Diagonal tension factor 7.4- Widths and load-carrying sections (V1-17/5) 7.4.1 Stiffener orientation 7.4.2 Frame orientation (V1-17/6) 7.5- Stresses in super-members 7.6- Stresses in super-stifffener (V1-17/8) 7.6.1 In skin 7.6.2 In stiffener 7.7- Stresses in frames (V1-17/9) 7.8- Allowable loads 7.8.1 At limit loads 7.8.2 At ultimate loads 7.9- Example (V1-17/11)

8 CALCULATING STIFFENED PANELS COMPRESSION AND BENDING 8.1- Introduction (V1-18-1) 8.2- Allowable stresses at ultimate load (V1-18-2) 8.3- Pocket folding load compression / bending 8.3.1 Detection of minimum folding stress cp 8.3.2 Search for load transmitted Pt 8.3.3 Example (V1-18-5) 8.4- Compression / bending stress (V1-18-8) 8.4.1 Position of extreme fibres 8.4.2 Compression / bending stresses at Max moment point 8.4.3 Compression / bending stresses at Min moment point (V1-18-9) 8.4.4 Limits, accuracy (V1-18-10) 8.4.5 Sign conventions 8.4.6 Example (V1-18-11) 8.5- Amplified bending margin (V1-18-14)


Contents

page •iii

Static stressing manual V1-1 • STIFFENED PANELS SYMBOLS USED A: Transverse Member pitch a: skin bay length B: Stiffener pitch b: Skin bay width C: Compression UL: Ultimate loads Cflf: Forced crippling coefficient LL: Limit loads CS: Single shear D: Fastener diameter E: Young's modulus Es: Secant modulus Et: Tangent modulus e: skin bay thickness et: Skin pad thickness es: Stiffener skin side flange thickness G: Coulomb's modulus GTDI: Effective shear modulus under TDi h: Stiffener height I: Section moment of inertia IG: General instability K: End fixity coefficient k: Diagonal tension factor L: Buckling effective length (column) Lt: Load-carrying width M: Bending moment MS: Static margin N: Normal force n: Stress-strain curve shape factor (Ramberg and Osgood) P: Compression force (P>0) PadmCE: Compression load-carrying capacity Pcrit: Column buckling load

p: Fastener pitch q: Transverse distributed load R: Loading rate Rc: Single compression loading rate RF: Reserve factor Rs: Single nominal shear loading rate r: Panel curvature radius S: Cross section area T: Shear load TD: Diagonal tension TDP: Pure diagonal tension TDI: Incomplete diagonal tension t: Shear flow W: Cross section static moment

α: Angle of diagonal tension ∆: Flattening deflection ε: Strain (expansion) Φ: Normal flow (tension or compression) γ: Shear strain η: Plasticity factor λ: Slenderness ratio v: Poisson factor σ: Normal stress σ0: Column buckling at zero slenderness ratio σ0.2: Conventional allowable compressive yield stress σcp: Skin bay buckling stress in compression σcrit: Column buckling σenr: Column buckling at zero slenderness ratio stiffener alone σfirp: Skin inter-rivet buckling stress σflf: Forced crippling stress σflr: Stiffener local buckling stress

Page V1-1/1 revision 1 © AEROSPATIALE 1998

Static stressing manual V1-1 • STIFFENED PANELS

σR: Failure tension stress τ: Shear stress τflf: Forced crippling shear loading stress τmax: Maximum shear stress (in Tresca line of thought) τR: Failure shear stress


Static stressing manual V1-1 • STIFFENED PANELS BIBLIOGRAPHY 1- Paul Kuhn, James P. Peterson, and L. Ross Levin: "A summary of diagonal tension", part 1 (methods and analysis), NACA TN 2661, Washington, May 1952.

2- Paul Kuhn, James P. Peterson, and L. Ross Levin: "A summary of diagonal tension", part 2 (experimental evidence), NACA TN 2662, Washington, May 1952.

3- Bruhn: "Analysis and design of flight vehicle structure" (in particular chapter C11).

4- Pablo Rodriguez: "Prédimensionnement d'une partie basse de fuselage en GLARE" (Presizing of a low part of the fuselage in GLARE), rapport de stage de fin d'étude (end of study report), INSA Toulouse, 1995.

5- Koos Verolme: "The development of a design tool for fibre metallic laminate compression panels", Ph. D. thesis, Delpht University, Nov. 1995.

6- F.R. Shanley: "Inelastic column theory", Journal of Aeronautical Sciences, Vol. 14, no. 5, pp 261-267, May 1947.

7- F. Engesser: "On the buckling strength of straight columns", Zeitschift für Architectur Ingenieurwesen, Vol. 35, no. 4, pp 455-462, 1889.

8- L. Euler: "Methodus inveniendi lineas curvas maximi minimive proprietate gaudentes", Annexe 1 ("De curvis elasticis"), Bousquet, Lausanne et Genève, 1744.



V1-1 •1-GENERAL INFORMATION A large proportion of aircraft structures is designed using a box beam type construction basis. The box beams are made of stiffened panels. As thin web beams are also considered as being part of this category, stiffened panels can be considered as the main basic element of aircraft airframes.

V1-1 •1.1-DEFINITIONS A stiffened panel is a flat or curved "skin" to which stiffeners are laid parallel to the normal dominating stress direction. For example, with box beams, this direction corresponds to the direction of internal forces generated by bending moments. The stiffeners are longitudinal members attached to the skin over their entire length either because they are "sewn" by means of bolts, rivets or tack welding and in this case they are called "fastened stiffeners", or because the whole skin is machined from a thick plate - in this case they are called "integrate stiffeners". Thus, stiffened panels comprise transverse members, the function of which is to provide supporting points, more or less evenly spaced, and more or less stiff for the stiffeners and possibly the skin. For example, frames perform this function in fuselage panels. Therefore, the skin is latticed in squares by members. The squares of panels defined in this manner are called "bays". For stressing purposes, the panels are broken down into "super-stiffeners" each consisting of a stiffener and half skin bays.

General information - Page V1-1•1/1 revision 1 © AEROSPATIALE 1998


Pocket bottom

BAY

Pad

Members

=

SUPER-STIFFENER B

=

= B =

Stiffener


Static stressing manual V1-1 • STIFFENED PANELS V1-1 •1.2-BEHAVIOUR, FAILURE MODES The skin of stiffened panels is load-carrying unlike the fabric in wooden and fabric constructions. These panels are capable of carrying/transferring two types of loads: - loads in the skin plane (Nx, Φy, Ty, My), - loads normal to the skin plane, for example due to the pressure applied to it (Tz).

Φy

Ty Φy

My

G Nx Tz

Stresses resulting from these forces are: - Normal (σx) and tangential (τxz ;τzx) in the stiffener, - Normal (σx), transverse (σy) and tangential (τxy ;τyx) in the skin. The crux of the problem in the stressing of stiffened panels is due to the fact that skin behaviour depends on the type and intensity of loading: • Tension: the static limit of skin strength at UL is given by the characteristics of the material. • Shear: this is also the case even though the folds occur beyond the critical shear stress of bays. These folds are oblique in relation to the edges. For this reason, through misuse of language, the term "diagonal tension" is used. • This property of dependence is further emphasised in the case of compression. This is related to the instability of thin sheets.


Static stressing manual V1-1 • STIFFENED PANELS Three types of instability are differentiated: • SKIN INSTABILITY: Thin sheets fold under relatively low compression and/or shear loads. In compression, the skin is saturated when its buckling stress is reached. At greater values, the load is picked up by the stiffeners and small portions of neighbouring skin which justifies the idea of breaking down panels into super-stiffeners. This saturation phenomenon does not exist under shear stressing. After folds occur, the skin transfers the excess load partly in the form of shear stress and the remainder as diagonal tension. The tangential flows at skin bay edges at constant loads are the same with or without folds. On the other hand, the diagonal tension induces normal flow at these same edges which causes overloads, especially in the stiffeners. However, these phenomena only create a dependence of the overall behaviour of stiffened panels on the intensity of the loads applied to them. Instability of the skin does not cause panel failure and therefore is not a cause of limitation at UL. • INSTABILITY OF SUPER-STIFFENERS: If the transverse members spaced at a pitch A are sufficiently stiff, the super-stiffeners behave as columns with a length A, generally simply supported. A

A

Therefore, the initial failure occurs in the most critical super-stiffener either by buckling of the column or by local buckling. Our panels are designed in this manner. • GENERAL INSTABILITY: This may occur if the wave nodes of the buckle pattern are no longer at transverse members due to insufficient stiffness of these members. High-speed amplification of this type of buckle pattern entails the total ruin and practically the explosion of the whole panel. This is to be prohibited. General information - Page V1-1•1/4 revision 1 © AEROSPATIALE 1998

Static stressing manual V1-1 • STIFFENED PANELS V1-1 •1.3-CALCULATION METHODOLOGY The information that this chapter of the manual makes possible to calculate is: - The stresses in the studied structural element corresponding to the applied loads (these values, amongst others, are needed as they can be compared with measurements made during non-destructive testing). - The load causing the occurrence of folds in skin bays. This should not occur too early (in general, not before 80% of LL). - The static failure load, i.e. the load allowable at UL. - The safety margin at UL, which is determined by comparing the allowable load and the applied load. In simple cases (stable shear and tension, tension), the calculation of the stresses under a given load does not cause any special problems. The panels are broken down into super-stiffeners that are assimilated to beams. The failure load is the one at which the equivalent stress on Von Mises reaches the allowable tension failure stress of the material (therefore, this criterion is systematically verified). If the skin folds before the failure load, then the stresses are no longer a simple linear function of loads. In this case, the general procedure to use to calculate a super-stiffener is as follows: - Select any stress value in the skin at the stiffener. - Calculate the "load-carrying section" corresponding to it. - The load corresponding to the selected skin stress is determined by summing the stresses on the load-carrying section. Also, the properties of the load-carrying section and the material properties are used to determine whether or not the failure load is reached at this time. By varying the initial parameters, a stress-load curve of the super-stiffener is obtained and we can therefore associate a stress level to with a given load. Naturally, this curve is increasingly monotonic. Data varying with the skin stress is required to determine the allowable load at UL. This problem is solved by plotting the evolution of the failure load (it is written in italics because the true failure load is naturally a constant) depending on this same initial parameter. This curve is decreasingly monotonic. The allowable load searched for is at the intersection of the load-stress and failure load-stress curves. Often, internal loads in the studied piece of structure are linear, with the general loads applied into the entire structure (more or less). For this reason, the margins are calculated by comparing the loads and not the stresses. General information - Page V1-1•1/5 revision 1 © AEROSPATIALE 1998

Static stressing manual V1-1 • STIFFENED PANELS V1-1 •1.4-FIELD OF APPLICATION The methods described in this chapter exclusively apply to the calculation of "typical areas": - Stiffeners are parallel one with the other and perpendicular to the transverse members. - The studied stiffener does not edge an opening. - No stiffener run-outs in the studied area. - Each rectangular skin bay may be considered as having a constant thickness. - Stiffeners may be considered as having constant cross sections between two transverse members. - The skin curvature is in the plane normal to the stiffeners. The special cases indicated below are discussed in Chapter V1-4 (thin web beams): - In plane bending of stiffeners edging an opening under the effect of the diagonal tension. - Other effects induced by the presence of an opening in the skin. - Secondary bending at run-outs. - Calculation of the fasteners binding the panels together. - Sizing of the fasteners binding the stiffeners to the skin.



V1-1 •2-PREPARATION OF BASIC DATA V1-1 •2.1-MATERIAL PROPERTIES STABILITY CALCULATIONS: The elastic-plastic behaviour of materials has to be taken into account in these calculations. To this end, the Ramberg and Osgood model (refer to V2-2 "Material Behaviour") is used: σ æ σ ö ε = + 0,002 ç ÷ è σ 0, 2 ø E Es =

n

σ ε

1 n 1− n = + Et Es E

Therefore, for each material used, the following is required: • Young's compression modulus, E. • Conventional allowable compression yield stress, σ0.2. • R.&O. factor, n.

ALL CALCULATIONS: Well designed stiffened panels (general buckling impossible) are damage tolerant structures. Choose "B values" type statistical properties.

Preparation of basic data - Page V1-1•2/1 revision 1 © AEROSPATIALE 1998

Static stressing manual V1-1 • STIFFENED PANELS V1-1 •2.2-SUPER-STIFFENER CROSS SECTIONS Assumption related to the section of curved panels: the curvature of panels is sufficiently small for them to be considered as flat panels.

CROSS SECTION (INTEGRAL STIFFENER):

Lt1

Lt2

d e1

et

e2

G

z

Initial cross section: Lti = Bi/2 B1

B2

PROPERTIES: Area: Moment of inertia/Gy: Centre of gravity offset: Load-carrying width: Bay thicknesses:

S I d Lt1, Lt2 e1, e2


Static stressing manual V1-1 • STIFFENED PANELS CROSS SECTION (FASTENED STIFFENER): Lt1

Lt2

Gp

Y d

G

e1

Y

dp e2

Gr

dr

z

Initial cross section: Lti = bi/2

b1

b2

b1 and b2 are the distances between the end rows of fasteners SKIN PROPERTIES: Cross section area: Static moment/YY: Moment of inertia/YY: Load-carrying width: Bay thickness: Centre of gravity offset:

STIFFENER PROPERTIES: Sp W YYp IYYp Lt1, Lt2 e1, e2 W YYp dp = Sp

Sr W YYr IYYr

dr =

W YYr Sr

Remark: W YYp is negative whereas W YYr is positive.


Static stressing manual V1-1 • STIFFENED PANELS SKIN + FASTENED STIFFENER PROPERTIES: The skin and the stiffener may be made from different materials. In this case, corrected cross sections are used so as to be in the simple case of a beam made of a homogeneous fictitious material (somewhat like an average material). The secant modulus and the tangent modulus of the equivalent homogeneous material is: Sp Sr Es = æç ö÷ Esp + æç ö÷ Esr ;S = Sp + Sr è Sø è Sø Sp Sr Et = æç ö÷ Etp + æç ö÷ Etr è Sø è Sø

The properties of the corrected section are: Esp Esr S = Sp + Sr = æç ö÷ Sp + æç ö÷ Sr è Es ø è Es ø Esp Esr W YY = æç ö÷ W YYp + æç ö÷ W YYr è Es ø è Es ø W YY d= S Esp Esr IYY = æç ö÷ IYYp + æç ö÷ IYYr è Es ø è Es ø I = IYY - Sd2

The secant modulus may be replaced by the Young's modulus in the following formulas as long as one remains in the linear elastic domain. The average stress calculated by using the properties of the corrected cross section σ is fictitious. The real stresses in the skin and the stiffener are: Esp σp = æç ö÷ σ è Es ø Esr σr = æç ö÷ σ è Es ø


Static stressing manual V1-1 • STIFFENED PANELS V1-1 •2.3-EXAMPLE The stress in the super-stiffener below, subject to a sufficiently low compression load for no instability to occur (stability calculations are discussed in paragraph 3) has to be calculated. The skin material properties (2024 PLT 351) are: Ep = 70300MPa, σ0.2p = 270 MPa, np = 7,05 The stiffener material properties (7075 T 73510) are: Er = 73800MPa, σ0.2r = 420 MPa, nr = 13,83

170 =

= 3

2 35 =

=

25 3 30

2 4 16

Stiffener:

Skin:

Sr = 185 mm²

Sxp0 = 375 mm²

IYYr = 62186 mm4

IYYp0 = 1485 mm4

dr = 13,9 mm

dp0 = - 1,86 mm

WYYr = 2572 mm3

WYYp0 = - 698 mm3


Static stressing manual V1-1 • STIFFENED PANELS Skin + stiffener section: S = 185 + 375 = 560 mm² HOMOGENEOUS FICTITIOUS MATERIAL: Young's modulus (linear elastic material assumption): 185 ö æ 375 ö 70300 = 71456 MPa E = æç ÷ 73800 + ç ÷ è 560 ø è 560 ø CORRECTED CROSS SECTION PROPERTIES: Skin + stiffener corrected cross section area: S = 560 mm² 73800 ö æ 70300 ö 375 = 560) (Check that: 185 + 375 =æç ÷ 185 + ç ÷ è 71456 ø è 71456 ø Static moment in relation to YY: 70300 ö æ 73800 ö 2571,5 = 1969,6 mm3 W YY = - æç ÷ 697,5 + ç ÷ è 71456 ø è 71456 ø Centre of gravity offset in relation to YY: 1969,6 d= = 3,52 mm 560 Moment of inertia in relation to YY: 70300 ö æ 73800 ö 62186 = 65687 mm4 IYY = æç ÷ 1485 + ç ÷ è 71456 ø è 71456 ø Moment of inertia at the centre of gravity: I = 65687 - 560 x 3,52² = 58759 mm4 EXAMPLE OF STRESS CALCULATION: If Nx is - 10000N, the "average stress" is: 10000 σ== - 17,9 MPa 560 The skin stress is: 70300 ö σp = - æç ÷ 17,9 = - 17,6 MPa è 71456 ø And the stiffener stress: 73800 ö σr = - æç ÷ 17,9 = - 18,1 MPa è 71456 ø


Static stressing manual V1-1 • STIFFENED PANELS V1-1 •3-SKIN BAY BUCKLING STRESSES V1-1 •3.1-ASSUMPTIONS Paragraph 2.2 explained that the properties of the cross section of a super-stiffener are determined assuming that the skin is flat. This assumption is sound as long as the curvature radius of our panels is long in comparison to the dimension of the skin bays (make sure that the proportions comply with information in paragraph 6.2). On the other hand, the curvature, no matter how slight it may be, has a non-negligible effect on bay folding stresses. In particular, a curved bay is more stable than a flat bay, especially under compression. Each bay is assimilated to a constant thickness rectangular plate, whether a flat or curved. The folding stresses σcp and τcp are buckling stresses (as indicated in paragraph 1.2 only shear and compression are taken into account). Generally, folding occurs at the same time on a set of contiguous bays. The appearance of the skin is then wavy under compression:

BUMP

HOLLOW

B

HOLLOW

BUMP

B

A

Stiffeners are stopped from rotating in their longitudinal axis by means of cleats located at the transverse members. However, members generally have an open section and their torsional stiffness is low. This explains the conservative boundary conditions indicated in paragraph 3.2. Refer to the next chapter (V1-2•BUCKLING OF THIN PLATES AND SHELLS) to make these calculations.

Skin bay buckling stresses - Page V1-1•3/1 revision 2 © AEROSPATIALE 1998

Static stressing manual V1-1 • STIFFENED PANELS V1-1 •3.2-BOUNDARY CONDITIONS If the members are made of open sections, the edging conditions to be selected to calculate pocket buckling stresses are: four simply hinged edges. • BAY WIDTH, b (measured in transverse member direction): b' (distance between fasteners) et e b"

B

Integrated stiffeners: Fastened stiffeners: All types if (et ≥ 3e):

b=B b = b' b = b"

• BAY LENGTH, a (measured in stiffener direction): Same principle with A, a', a".

V1-1 •3.3-DESIGN Generally, requirements stipulate that bay buckling shall not occur before a certain percentage of limit load. To check out this condition, the folding stresses are compared with the average stresses at LL (linear static).


Static stressing manual V1-1 • STIFFENED PANELS V1-1 •3.4-EXAMPLE • EXAMPLE 1: Use the example in paragraph 2.3, assuming the bays are flat, under the average stresses at LL: σ = 45MPa ;τ = 50 MPa

Skin material: 2024PLT3 Ep = 70300 MPa ;ν = 0,33 Bay geometry to the right and to the left of the stiffener (zero curvature): B = b' = b = 170 mm ; A = a' = a = 530 mm ; e = 2 mm Boundary conditions: Four simply hinged edges. SINGLE COMPRESSION æ a ≥ 1ö Þ kc = 4 ç ÷ ( ) èb ø

Refer to example in V1-2.2.5: σcp0 = 36 MPa ; cR=

σ = 1,25 σcp0

SINGLE SHEAR æ b = 0,32ö Þ ks = 5,74 ç ÷ ( ) èa ø

Refer to example in V1-2.2.5: τcp0 = 52 MPa ; sR=

τ τcp0

= 0,962

COMPRESSION-SHEAR INTERACTION

æ Rc + çR = σ = τ = ç σcp τcp è

R 2c + 4 R s2 2

ö = 1,77÷ Þ folding at 56% of LL ÷ ø


Static stressing manual V1-1 • STIFFENED PANELS • EXAMPLE 2: Use example 1 assuming that there are curved bays (r=2820mm), under average stresses at LL: σ = 45MPa ;τ = 50 MPa SINGLE COMPRESSION

See the example in V1-2.3.5: σcp0 = 45 MPa ; cR=

σ =1 σcp0

SINGLE SHEAR

See the example in V1-2.3.5: τcp0 = 57 MPa ; sR=

τ τcp0

= 0,877

COMPRESSION-SHEAR INTERACTION

æ Rc + çR = σ = τ = ç σcp τcp è

R 2c + 4 R s2 2

ö = 1,51÷ Þ folding at 66% of LL ÷ ø


Static stressing manual V1-1 • STIFFENED PANELS V1-1 •4.6-CALCULATION OF THE ALLOWABLE LOAD AT UL USING THE JOHNSON FORMULA V1-1 •4.6.1-Principle The Johnson formula is used to estimate the critical buckling stress of a column with a low slenderness ratio using a smoothing curve. This curve is a parabola expressing σcrit as a function of the slenderness ratio between λ = 0 and the limit value λ' 0. The point of tangency with the Euler curve corresponds to this limit value. σcrit is written as follows: λ' 0 =∏

2E σ0

λ2 æ ö Johnson: (0≤ λ ≤ λ' σ 20 ÷ 0) Þ ç σcrit = σ 0 − 2 è ø 4∏ E ∏ ²E ö æ (λ ≥ λ' ÷ 0) Þ ç σcrit = è λ² ø

Euler:

σ0 is the allowable stress at zero slenderness ratio in the super-stiffener (refer to paragraph 4.4.3). σcrit σ0

Johnson σ0/2 Euler

λ' 0

λ

V1-1 •4.6.2-Limits, accuracy This method is used for the stressing of panels made of aluminium alloy, typically used in the aeronautical industry. Stiffened panels under single compression - Page V1-1•4/15 revision 2 © AEROSPATIALE 1998

Static stressing manual V1-1 • STIFFENED PANELS The accuracy of the failure load calculation depends on the form of the curves (σ, ε), the quality of the local buckling analysis and the stiffener warping analysis. The result is satisfactory (skin in 2024, stiffeners in 7xxx) to slightly conservative (skin and stiffeners in 7xxx). However, the stability of a compressed structural element may be affected by imperfections. For sizing, it is advisable to take into account minimum 15% margins, confirmed by partial tests. Moreover, these margins may be used to readjustσ0.

V1-1 •4.6.3-Methodology Now the following data is available:

( Ep ;σ 0.2 p

• Basic material data (refer to paragraph 2.1):

n;p ) ( E, r σ ; 0 . 2 r nr ) ;

• Basic geometrical data (refer to paragraphs 2.2 and 4.3): (S 0 ;I 0 L;) • Pocket folding (refer to paragraph 3):

(σ

cp1

;σ cp 2

• Local buckling stresses (refer to paragraph 4.4.1):

(σ

firp

;σ flr

• Zero slenderness ratio allowable stress (refer to paragraph 4.4.3):

(σ

0p

)

)

;σ 0 r σ;0

)

Also, a set of rules and equations can now be used to calculate:

• The load-carrying section area for a given skin stress (refer to paragraph 4.2): (Sp, S) • ε, σr, and the tangent and secant moduli corresponding to σp, using the material properties: æ σ ö σ ε = r + 0.002 ç r ÷ Er è σ 0.2 r ø

Esr =

nr

=

σp Ep

æ σp ö ÷÷ + 0.002 çç è σ 0 .2 p ø

np

σp σr ;E sp = ε ε

• The secant modulus of the equivalent homogeneous material corresponding to σp (refer to paragraph 2.2): æ Sp ö æ Sp ö æS ö æS ö Es = ç ÷ Esp + ç r ÷ Esr ;σ = Esε = ç ÷ σp + ç r ÷ σr è Sø è Sø è Sø è Sø

• The inertia of the load-carrying section (refer to paragraphs 2.2 and 4.2): I Stiffened panels under single compression - Page V1-1•4/16 revision 2 © AEROSPATIALE 1998

Static stressing manual V1-1 • STIFFENED PANELS • Slenderness ratio: L I λ = ;ρ = S ρ By varying the parameter σp:

• The curve giving λ in function of σ or vice-versa (this equation is biunivocal) is obtained. • Also, the curve giving the buckling load of the super-stiffener as a function of λ and vice-versa (this equation is also biunivocal) is obtained: λ' 0 =∏

2E σ0

æ ö λ2 Johnson: (0≤ λ ≤ λ' σ 20 ÷ 0) Þ ç σ crit = σ 0 − 2 è ø 4∏ E æ ∏2 Eö = ) Þ σ Euler: (λ ≥ λ' 0 ç crit ÷ è λ2 ø Buckling of the super-stiffener occurs when:

σ = σcrit Pcrit = Sσcrit The load-carrying capacity of the super-stiffener is the allowable load at UL. If the stiffener or skin local buckling value is reached before Pcrit then the load-carrying capacity is the value of the corresponding P:

PadmUL = min {P crit, P(σfirp), P(σflr)}

Stiffened panels under single compression - Page V1-1•4/17 revision 2 © AEROSPATIALE 1998

Static stressing manual V1-1 • STIFFENED PANELS V1-1 •4.6.4-Example Use the example in paragraph 4.4.5. Allowable stress at zero slenderness ratio of the super-stiffener

σ0p = 267 MPa ;ε0 = 0.565% σ0r = 380 MPa Effective width and load-carrying section as per paragraph 4.2, for σp = σ0p: 36 Lt = 85 = 31.2 ;∆S = 2 x (85 - 31.2) x 2 = 215 267 Sp = Sp0 - ∆S = 160 ; S =p S+ Sr = 345 Average allowable stress at zero slenderness ratio (refer to paragraph 4.4.3): æ Sp ö æS ö σ0 = ç ÷ σ0p + ç r ÷ σ0r ≈ 328 MPa è Sø è Sø

Calculation of the load-carrying capacity and stresses at UL:

A spread sheet was used for these calculations (tables and graphs on following pages). Stresses at UL: P = 38000 N → (σp = 84 MPa ;σr = 88 MPa) Load-carrying capacity: Pcrit = 89800 N → (σp = 225 MPa ;σr = 277 MPa)


Static stressing manual V1-1 • STIFFENED PANELS Modelling of material curves:

epsilonp 0.00051 0.00114 0.00217 0.00268 0.00309 0.00333 0.0036 0.00376 0.00429 0.00472 0.00523 0.00565 0.00584

sigmap 36 80 150 180 200 210 220 225.154 240 250 260 267.05 270

epsilonr 0.0005 0.00114 0.00215 0.00267 0.00308 0.00332 0.00359 0.00376 0.00428 0.00471 0.00521 0.00565 0.00583 0.00585

sigmar 37 84 159 197 227 245 265 276.916 313 340 364 380 385.5 386

Epsilon 0.051% 0.114% 0.217% 0.268% 0.309% 0.333% 0.360% 0.376% 0.429% 0.472% 0.523% 0.565% 0.584%

Sigmap 36.0 80.0 150.0 180.0 200.0 210.0 220.0 225.2 240.0 250.0 260.0 267.1 270.0

Sigmar 37.8 84.0 159.8 197.4 227.7 245.5 265.5 276.9 313.6 340.2 364.8 380.0 385.7

Sigma 36.6 81.7 154.7 188.7 214.1 228.2 243.6 252.1 278.7 297.8 316.0 327.6 332.2

Input data is the skin and stiffener stresses, columns 2 and 4. The stiffener stresses corresponding to (ε=εp) in column 7 are obtained by linear interpolation in columns 3 and 4. np

σp

æ σp ö æ σ ö σ P ÷÷ ;εr = r + 0.002 ç r ÷ ;σ = + 0.002 çç εp = Ep Er S è σ 0.2 r ø è σ 0 .2 p ø nr

Secant modulus:

epsilonp 0.051% 0.114% 0.217% 0.268% 0.309% 0.333% 0.360% 0.376% 0.429% 0.472% 0.523% 0.565% 0.584%

Sp 375.0 263.1 201.6 187.1 179.2 175.8 172.5 171.0 166.7 164.0 161.5 159.8 159.2

S 560.0 448.1 386.6 372.1 364.2 360.8 357.5 356.0 351.7 349.0 346.5 344.8 344.2

P (N) 20492 36588 59799 70193 77978 82323 87081 89720 98025 103941 109485 112981 114319

Esp 70300 70277 69270 67286 64808 63115 61086 59905 56000 52981 49702 47268 46227

Esr 73800 73800 73800 73798 73790 73773 73727 73676 73180 72095 69739 67258 66031

æ σ cp ö σ ÷ ; P = pSσp + Srσr ; E Sp = Sp0 - be çç1− s= ÷ E σp ø è


Es 71456 71731 71438 70524 69370 68580 67627 67062 65038 63113 60399 57993 56873

Static stressing manual V1-1 • STIFFENED PANELS Curves (σ, ε) of equivalent homogeneous skin and stiffener materials:

400 350 Stress (MPa)

300

Sigmap Sigmar Sigma

250 200 150 100 50 0 0

0.002

0.004

0.006

Epsilon

Critical load and stress:

Sigma 37 86 155 189 221 228 244 252 279 298

S

I 560 442 387 372 362 361 358 356 352 349

Lambda sigcrit 58759 54058 51071 50436 50263 50267 50311 50350 50481 50533

51.74 47.95 46.11 45.52 45.01 44.90 44.68 44.56 44.24 44.05

226 240 247 249 251 251 252 252 253 254

Sigmap Sigmar P 36 84 150 180 205 210 220 225 240 250

38 88 160 197 236 245 266 277 314 340

Pcrit 20492 38002

89786

89788

Columns 3 and 4: ö E 1 æ E sp e (σ = 37)→ S0 ;∆S = S - S0 ; WYYp = WYYp0 + æç e t − ö÷ ∆S ; d = ç W YYp + sr W YYr ÷ è S è Es Es 2ø ø 2 E sp æ E S e e2 ö IYYp = IYYp0 - ç æç e t − ö÷ + ÷ ∆S ; I = IYYp + sr IYYr - Sd2 ;λ = L Es I 2ø 12 ø Es èè

Column 5: σcrit =σ0 -

λ2 σ 20 2 4∏ E


Static stressing manual V1-1 • STIFFENED PANELS Curves (σ, λ) and (σcrit, λ):

350 300 sigcrit

250

Sigma 200 150 100 50 43.00

44.00

45.00

46.00

47.00

48.00

Slenderness ratio


Static stressing manual V1-1 • STIFFENED PANELS V1-1 •4.7-SIMPLIFIED APPROACH TO THE JOHNSON FORMULA V1-1 •4.7.1-Method The method consists in using the diagram on the following page to obtain an estimated value of Pcrit. This is nothing else than a parametric bundle of Euler-Johnson curves. The parameter σ ö æ ç100 x 0 ÷ is dimensionless. Each curve of the diagram associates the dimensionless variable è Eø σ λ to crit . σ0 Proceed as follows: • Determine the Young's modulus of the equivalent homogeneous material and the allowable stress at zero slenderness ratio as previously explained (refer to paragraph 4.6.3). • Determine S, I and λ corresponding to σ0, i.e. σp0. • Determine σcrit using the diagram. This method may be used for sizing. For example, it makes it possible to quickly assess the effect of a change in material.

Column buckling in the transition domain 1.2 J o h n s o n

Sigmacrit/Sigma0

1 0.8 0.2 0.6 0.6

0.7

0.4

0.3 0.5

E u l e r

0.4

0.8 0.2 0

0.9 1

Each curve is defined for: 100*Sigma0/E=Cte

0

10

20

30

40

50 60 Lambda

70

80

90

100 110


Static stressing manual V1-1 • STIFFENED PANELS V1-1 •4.7.2-Example Use the examples in paragraphs 4.4.5 and 4.6.4: Allowable stress at zero slenderness ratio for the super-stiffener

σ0p = 267 MPa σ0r = 380 MPa

Effective width and load-carrying section as per paragraph 4.2, for σp = σ0p: Sp = 159 ; S =p S+ Sr = 345 mm² Average allowable stress at zero slenderness ratio (refer to paragraph 4.4.3): σ0 = 328 MPa Young's modulus of the equivalent homogeneous material:

E = 71456 MPa

Calculation of the column buckling load: σ0 328 = = 0.46% σ → crit = 0.77 E 71456 σ0 λ ≈ 44

σcrit = 0.77 x 328 = 252.6 MPa Pcrit = 345 x 252.6 = 87100N



PAGE INTENTIONALLY LEFT BLANK



V1-1 •4-CALCULATION OF STIFFENED PANELS UNDER SINGLE COMPRESSION

G y

P z

A super-stiffener subject to a compression load P=-Nx, behaves like a column with a cross section depending on P. At a given P, the section is assumed constant along the longitudinal centreline of the stiffener. The instability mode of super-stiffeners is described in paragraph 1.2 ("Behaviour, failure modes"). The transverse members are of sufficient stiffness and therefore general buckling of the panel cannot occur. As seen in paragraph 1.3, the problem of the stressman is: • To establish the relationship between the applied load P and stresses σp and σr. • To calculate the failure load by column buckling, Pcrit and the allowable load at UL, PadmCE (these two loads are not necessarily merged).


Static stressing manual V1-1 • STIFFENED PANELS V1-1 •4.1-PRINCIPLES • LOAD-CARRYING WIDTH PRINCIPLE: As long as the skin stress σp is less than the smallest buckling stresses in compression of the pocket bottoms involved, σp is uniform. The pockets saturate after folding. The average stress in their centres is approximately constant whereas the stress continues to increase at the stiffener at the location where the skin is fastened to the stiffener. The load-carrying width concept is used to simplify calculations. These widths are fictitious widths of skin such that, assuming σp is uniform, the right distribution of loading in the stiffener and in the skin is found. σ

Lt1

Lt2

σp σcp2

σcp1

Lt1.0

Lt2.0

The 0 index relates to the initial cross section, i.e. the section before folding. The loadcarrying widths of the initial section are given in paragraph 2.2 ("Super-stiffener sections"). The "load-carrying cross section" deduced from the load-carrying widths therefore depend on P, σcp1, and σcp2. In fact, below it is explained that it is easier to deduce P from σp than the opposite.

• FAILURE LOAD: The cause of failure of the super-stiffener is either local buckling or column buckling.


Static stressing manual V1-1 • STIFFENED PANELS In this text, local buckling designates highly localised phenomena, i.e. with a small buckle pattern wave length, like local buckling of the inner flange or the web of the stiffener (if it is of the open section type) or inter-rivet buckling of the skin. The phenomena occur abruptly and cause just an abrupt drop in the strength characteristics of the column.Therefore, local buckling stress constitutes peak values. There are other stiffener buckling phenomena that could be called "semi-local". Warping of open section stiffeners, or "crippling". It occurs more progressively and have a larger wave length and entail permanent deformation of transverse sections. If this type of phenomenon can occur before the yield stress of the material, it is combined with column buckling. The result is the same as if σ0.2 was reduced. As σ0.2, the warping stress and the crippling stress are parameters affecting column buckling and not simple peak values. For as long as the slenderness ratio λ of a column is sufficient, the stress and the critical column buckling load are given by the Euler formula: σcrit =

L I ∏2 E ;λ = ;ρ = 2 λ S ρ

Pcrit =

∏ 2 EI L2

This formula is only valid in the linear elastic domain, being for σcrit less than the material limit of proportionality. However, in aircraft construction, small slenderness ratios are used. Buckling of super-stiffeners, when it can occur appears in the elastic-plastic domain. This explains the two problem solving methods described in this chapter, based on: - The Engesser formula. - The Johnson formula.

V1-1 •4.2-EFFECTIVE WIDTH AND LOAD-CARRYING CROSS SECTIONS The load-carrying widths to the right and to the left of the stiffener correspond to a given skin stress σp, and are calculated using the Karman formula (see figures in paragraph 2.2):



(σp ≤ σcpi)

Þ ( Lti = Lti,0) ; Lti,0 =

(σp ≥ σcpi)

æ Þ ç Lti = Lti.0 è

bi 2

σcpi ö ÷ σp ø

To calculate the characteristics of the load-carrying section, the simplest method is to subtract: rather than reconstructing the entire cross section using load-carrying widths, remove from the initial section (index 0) the two small skin rectangles that are not "load-carrying". Then the following is obtained: 2

∆Si = ( Lti,0 − Lti ) ei ;∆S = å ∆Si i =1

S = S0 - ∆S

INTEGRATED STIFFENER COLUMN:

d = d0 +

1 S

å æçè d 2

i =1

0

−

ei ö ÷ ∆Si 2ø

2 2 æ e2 ö ei ∆I = å ç æç d 0 − ö÷ + i ÷ ∆Si è 2ø 12 ø i =1 è

I = I0 - ∆I - (d 0 − d ) S 2

INTEGRATED STIFFENER COLUMN: 2 ei W YYp = WYYp0 + å æç et − ö÷ ∆Si è 2ø i =1 2 ææ e 2i ö ei ö IYYp = IYYp0 - å ç ç et − ÷ + ÷ ∆Si è 2ø 12 ø i =1 è 2

Naturally, the properties of the stiffener do not change. To calculate the properties of the super-stiffener, the corrected section design rules given in paragraph 2.2 are used.

V1-1 •4.3-EFFECTIVE LENGTH OF BUCKLING, END FIXITY COEFFICIENT The true length of a super-stiffener A is equal to the distance between the transverse members delimiting it.


Static stressing manual V1-1 • STIFFENED PANELS The effective length of buckling, L, is the theoretical length of the column of the same cross section both ends simply supported, buckling under the same critical load. L therefore depends on the supporting conditions at the ends of the super-stiffener: L = KA

The main fixity end coefficients, K, are given below. For more details refer to Chapter V2-5 "COLUMNS". In current areas, use K=1 (beam with two simple support ends). MAIN END FIXITY COEFFICIENTS:

P

P

L

K=0,5

K=1

P

P

K=0,7

K=2


Static stressing manual V1-1 • STIFFENED PANELS V1-1 •4.4-ALLOWABLE STRESSES AT ULTIMATE LOAD (UL) V1-1 •4.4.1-Local buckling SKIN: Inter-rivet buckling, σfirp. STIFFENER: σflr is the smallest local buckling stress of the stiffener (web, inner flange, inter-rivets, etc.)

• Inter-rivet buckling: The pitch between the fasteners binding the stiffener to the skin, the thickness of the skin pad under the stiffener and the thickness of the stiffener skin side flange must be sized so that inter-rivet buckling cannot occur. Skin and stiffener inter-rivet buckling stresses σfirp and σfirr are calculated considering that the length of the strips of sheet are equal to the fastener pitch p and that the non-loaded edges are free. The end fixity coefficient K depends on the type of link: FLAT HEAD SCREW, TACK WELDS: K=0.54 COUNTERSUNK HEAD SCREW, RIVETS: K=0.66

p

Refer to Chapter V1-3 "STIFFENERS" to calculate the inter-rivet buckling stresses.

• Other local buckling stresses: Refer to Chapter V1-3 ("STIFFENERS").


Static stressing manual V1-1 • STIFFENED PANELS V1-1 •4.4.2-Crippling, warping, allowable stress at zero slenderness ratio in the stiffener Refer to Chapter V1-3 ("Stiffeners") to calculate the crippling σcr and the warping σdr stress. The allowable stress associated with a zero slenderness ratio in the stiffener σenr, is the smallest value amongst the conventional allowable compression yield stress of the material, warping stress and crippling stress:

σend = min {σ 0.2 r ;σcr ; σdr}

V1-1 •4.4.3-Allowable stress at zero slenderness ratio in the super-stiffener This is the stress above which the deformation of the transverse cross section is permanent. This limit is considered to be reached when the skin stress in σ0.2 or when the stiffener stress is σenr. EXAMPLE: Stiffener Equivalent homogeneous material

σenr = σ0r σ0.2p

σ0

Skin

σ0p

ε0r

ε0p

In this illustration, the limit is due to the stiffener even though the zero slenderness ratio allowable stress in the stiffener is greater than the allowable compressive yield stress of the skin.


Static stressing manual V1-1 • STIFFENED PANELS V1-1 •4.5-CALCULATION OF THE ALLOWABLE LOAD AT UL USING THE ENGESSER FORMULA V1-1 •4.5.1-Principle This is the Euler formula in which the Young's modulus, E, is replaced by the tangent modulus Et: σcrit =

∏ 2 Et ∏ 2 EtI ; P = crit λ2 L2

However, when applied as such, these equations do not make it possible to take into account possible warping or crippling of the stiffener.

V1-1 •4.5.2-Modified Engesser formula In fact, this is a modification of the laws of behaviour of materials: Use the Ramberg and Osgood model (refer to paragraph 2.1) replacing σ0.2 by σ0: σ σ σ 1 n 1− n + 0.002 æç ö÷ ; E ; = + s= è σ0 ø E ε Et Es E n

ε=

Furthermore, whatever the slenderness of the column: max { σcrit} = σ0

V1-1 •4.5.3-Limits, accuracy This method is used to calculate panels built of any homogeneous and isotropic metallic material of which the increasing monotonic curves (σ, ε) can be modelled by means of R&O formulas. The accuracy of the failure load calculation is satisfactory as long as the possibilities of local buckling and warping of stiffeners is correctly analysed. However, stability of a compressed structural element may be affected by imperfections. For sizing, it is advisable to take into account 15% minimum margins confirmed by partial tests. Furthermore, these margins may be used to readjustσ0.

V1-1 •4.5.4-Methodology Now the following data is available: Stiffened panels under single compression - Page V1-1•4/8 revision 2 © AEROSPATIALE 1998

Static stressing manual V1-1 • STIFFENED PANELS • Basic material data (refer to paragraph 2.1):

( Ep ;σ 0.2 p

n;p) (E, r σ ; 0 . 2 r nr ) ;

• Basic geometrical data (refer to paragraphs 2.2 and 4.3):

(S0 ;I 0 L;)

• Skin pocket folding (refer to paragraph 3):

(σcp1 ;σcp2)

• Local buckling stresses (refer to paragraph 4.4.1):

(σfirp ;σflr )

• Zero slenderness ratio allowable stresses (refer to paragraph 4.4.3):

(σ 0 p ;σ 0 r )

Also, a set of rules and equations can now be used to calculate:

• The load-carrying section area for a given skin stress (refer to § 4.2): (Sp, S) • ε, σr and the tangent and secant moduli corresponding to σp, using the modified material properties:

σr σr + 0.002 æç ö÷ ε= è σ0r ø Er

nr

σp æ σp ö ε= + 0.002 ç ÷ è σ0p ø Ep

np

;E sr =

σr 1 nr 1 − nr ; = + ε Etr Esr Er

;E sp =

σp 1 np 1− np ; = + ε Etp Esp Ep

• The secant modulus and the tangent modulus of the equivalent homogeneous material corresponding to σp (refer to § 2.2): Sp Sr æ Sp ö E + æ Sr ö E Es = æç ö÷ Esp + æç ö÷ Esr ; E ÷ ç ÷ t =ç è Sø è Sø è S ø tp è S ø tr

• The inertia of the load-carrying section (refer to § 2.2 and 4.2): I By varying the parameter σp:

• The curve giving P as a function of σp, σr or σ or vice-versa (this equation is biunivocal) is obtained: P = Spσp + Srσr = Sσ • Also, the curve giving the buckling load of the super-stiffener as a function of σp, σr or σ and vice-versa (this equation is also biunivocal) is obtained:


Static stressing manual V1-1 • STIFFENED PANELS Pcrit =

∏ 2 EtI L2

Column buckling of the super-stiffener occurs when:

P = Pcrit ; and:σcrit =

Pcrit ≤ σ0 S

The load-carrying capacity of the super-stiffener is the allowable load at UL. If the stiffener or skin local buckling value is reached before Pcrit then the load-carrying capacity is the value of the corresponding P:

PadmCE = min {Pcrit, P( σfirp), P( σflr )}

V1-1 •4.5.5-Example Use the example in paragraph 2.3 The material properties of the skin (2024 PLT 351) are: Ep = 70300MPa, σ0.2p = 270 MPa, np = 7.05 The material properties of the stiffener are (7075 T 73510) are: Er = 73800MPa, σ0.2r = 420MPa, nr = 13,83 The load applied at UL is: P = 3800daN 170 =

= 3

2 35 =

=

25 3 30

2 4 16


Static stressing manual V1-1 • STIFFENED PANELS Stiffener: Sr = 185 mm² IYYr = 62186 mm4 dr = 13,9 mm WYYr = 2572 mm3

Skin: Sxp0 = 375 mm² IYYp0 = 1485 mm4 dp0 = 1,86 mm WYYp0 = - 698 mm3

Pocket folding

Refer to example 1 in paragraph 3.3 (a1 = a2 = 530 mm ; 1b= b2 = 170 mm): σcp1 = σcp2 = 36 MPa Inter-rivet skin buckling

Linked by countersunk head bolts, pitch 25 mm: et = 3 mm ; K = 0,66 Details of the skin local buckling stress calculation, as well as that of the other local buckling stresses, are given in Chapter V1-3 ("STIFFENERS"). σfirp = 289 MPa Stiffener local buckling

inner flange: 504 MPa web: 476 MPa inter-rivet: 413 MPa

σflr = 413 MPa

Allowable stress at zero slenderness ratio of the stiffener

Allowable compressive yield stress: Warping: Crippling:

σ0.2r = 420 MPa σdr = 380 MPa σcr = 461 MPa

380 380 ö + 0.002 æç σenr = 380 MPa ;ε0r = ÷ è 420 ø 73800

13.83

= 0.565%

Allowable stress at zero slenderness ratio of the super-stiffener Skin allowable compressive yield stress: σ0.2p = 270 MPa ; ε0p =

270 + 0.002 = 0.584% 70300

The limit is due to the stiffener:

(ε 0 p > ε 0 r )

np æ σ0 p æ σ 0 p ö ö÷ ç and (σ 0 r = 380 MPa ) Þ ç σ 0 p ε 0 = 0.565% = + 0 .002 ç ÷ è σ 0.2 p ø ÷ø Ep è


Static stressing manual V1-1 • STIFFENED PANELS This equation is not reversible. A numerical solution has to be found:

σ0p = 267 MPa checks:

0.565 267 267 ö + 0.002 æç = ÷ è 270 ø 100 70300

7. 05

Pocket folding load

(σp = σcp = 36 MPa ) → ( Sp = Sp 0 = 375 mm 2 and

36 36 ö + 0.002 æç ε= ÷ è 270 ø 70300 σr = 37.8 MPa

7. 05

≈

;S = S0 = 560 mm 2 )

36 = 0.512% 70300

Stress which checks out: 13.83 37.8 37.8 ö 37.8 æ + 0.002 ç = 0.512% ε= ≈ ÷ è 380 ø 73800 73800 The load corresponding to the stresses is: Pcp = 36 x 375 + 37.8 x 185 = 20492 N, i.e. 81% LL 38000 (The load applied at LL is equal to: ≈ 25300 N). 1.5

Calculation of the load-carrying capacity and the stresses at UL

It can be seen that the local buckling stresses are greater than the allowable stresses at a zero slenderness ratio both in the skin and in the stiffener. Therefore, the failure mode of this super-stiffener is column buckling. For these calculations, a spread sheet was used. Stresses at UL: P = 38000 N → (σp = 84 MPa ;σr = 88 MPa) Load carrying capacity: Pcrit = 90200 N → (σp = 225 MPa ;σr = 279 MPa)


Static stressing manual V1-1 • STIFFENED PANELS Modelling of material curves:

epsilonp 0.00051 0.0012 0.00217 0.00268 0.00323 0.00336 0.00364 0.00381 0.00436 0.00481 0.00536 0.0058 0.006

sigmap 36 84.13 150 180 205 210 220 225.38 240 250 260 267.05 270

epsilonr 0.00051 0.0012 0.00217 0.00268 0.00323 0.00335 0.00364 0.00381 0.00436 0.00481 0.00536 0.0058 0.006 0.00609

sigmar 37.8 88.4 160 198.1 237.9 247.1 267.5 279.2 311.9 332.2 349.3 359.5 363.4 365

Epsilon 0.051% 0.120% 0.217% 0.268% 0.323% 0.336% 0.364% 0.381% 0.436% 0.481% 0.536% 0.580% 0.600%

Sigmap 36.0 84.1 150.0 180.0 205.0 210.0 220.0 225.4 240.0 250.0 260.0 267.1 270.0

Sigmar 37.8 88.4 160.0 198.1 237.9 247.2 267.5 279.2 311.9 332.2 349.3 359.5 363.4

Sigma 36.6 85.9 154.8 189.0 221.8 229.1 244.6 253.4 277.8 293.6 307.7 316.7 320.2

Input data is the skin and stiffener stresses, columns 2 and 4. The stiffener stresses corresponding to (ε=εp) in column 7 are obtained by linear interpolation in columns 3 and 4. σp æ σp ö εp = + 0.002 ç ÷ è σ0p ø Ep

np

σr σr + 0.002 æç ö÷ ;εr = è Er σ0r ø

nr

;σ =

P S

Tangent and secant moduli: epsilon 0.051% 0.120% 0.217% 0.268% 0.323% 0.336% 0.364% 0.381% 0.436% 0.481% 0.536% 0.580% 0.600%

Sp 375.0 257.4 201.6 187.1 177.5 175.8 172.5 170.9 166.7 164.0 161.5 159.8 159.2

S 560.0 442.4 386.6 372.1 362.5 360.8 357.5 355.9 351.7 349.0 346.5 344.8 344.2

P(N) 20492 38003 59835 70318 80393 82648 87445 90168 97710 102468 106614 109196 110207

æ Columns 2 and 4: Sp = Sp0 - be ç1 − è

Esp 70300 70266 69187 67050 63536 62591 60436 59131 55082 51932 48536 46032 44966

Esr 73800 73800 73800 73793 73730 73684 73484 73254 71589 69014 65206 61972 60527

Es 71456 71744 71395 70403 68739 68279 67188 66472 63765 60987 57436 54584 53331

Etp 70300 70061 63141 52396 40159 37627 32689 30150 23848 20123 16894 14905 14139

Etr 73800 73800 73797 73753 73307 72992 71630 70116 60606 49568 38255 31464 28986

σcp ö ÷ ; P = pSσp + Srσr σp ø −1

Columns 5 to 7: Es =

Et 71456 71624 68241 63015 57077 55762 52838 50925 43185 35730 28299 23789 22120

σ æ n + 1 − n ö ; column 10: E= Sp E + Sr + E ; columns 8 and 9: E ÷ t =ç t tp tr è Es ε S S E ø


Static stressing manual V1-1 • STIFFENED PANELS Critical stresses and loads: Sigma 37 86 155 189 222 229 245 253 278

S WYYp -698 560 -462 442 -351 387 -322 372 -302 362 -299 361 -293 358 -289 356 -281 352

d 3.52 4.96 6.00 6.42 6.84 6.93 7.13 7.24 7.52

IYYp 1485 975 733 671 629 622 608 601 582

I 58759 54059 51088 50478 50333 50341 50383 50411 50435

P 20492 38003 59835 70318 80393 82648 87445 90168 97710

Pcrit Sigmap Sigmar 38 147524 36 136043 84 88 122492 111764 100941 98630 93536 279 90200 225 76526

Columns 3 to 6: e 1 Esp Esr (σ = 37) → S0 ;∆S = S - S0 ; WYYp = WYYp0 + æç et − ö÷ ∆S ; d = æç W YYp + W YYrö÷ è ø è ø 2 S Es Es 2 ææ Esp Esr eö e2 ö IYYp = IYYp0 - ç ç et − ÷ + ÷ ∆S ; I = IYYp + IYYr - Sd2 2ø 12 ø Es Es èè

Column 8: Pcrit =

∏ 2 EtI L2

P and Pcrit as a function of σ: P and Pcrit as a function of the average stress Sigma

160000 140000 120000 100000

P

80000 Pcrit

60000 40000 20000 0 0

100

200

300

400

Sigma (MPa) Stiffened panels under single compression - Page V1-1•4/14 revision 2 © AEROSPATIALE 1998


V1-1 •5-CALCULATION OF FLAT STIFFENED PANELS UNDER SINGLE NOMINAL SHEAR

Ty

G z

x

Flat stiffened panels designed to transfer shear in the skin plane are classified in three categories corresponding to their behaviour mode: - Panels under stable shear. - Panels under pure diagonal tension. - Panels under semi-diagonal tension or incomplete diagonal tension. There are no special comments to make concerning the stressing of stable panels in shear. This paragraph is dedicated to incomplete diagonal tension. The approach given here is based on theoretical analyses, readjusted following tests. it applies to the domain defined in the "Limits" paragraph.

Flat stiffened panels under shear - Page V1-1•5/1 revision 2 © AEROSPATIALE 1998

Static stressing manual V1-1 • STIFFENED PANELS V1-1 •5.1-DIAGONAL TENSION THEORIES Let us consider a rectangular plate, with thickness e, surrounded by members hinged together at their ends, subjected to a shear flow t. The members are infinitely stiff in bending and symmetrical in relation to the centre plane of the skin. The horizontal members have the same cross section, as do the vertical members. y

t

t x B e

A

The stable shear behaviour is as follows at the scale of a 2D-element from the skin: σt

τ CS

=

Π/4 σc σc = - σt = - τ

At and after a certain load level, folds appear caused by the diagonal compression stress. The excess load then causes the augmentation of the diagonal tension stress alone. The pure diagonal tension theory is from Wagner: If the mechanic and geometric properties of the skin and the load are such that the diagonal compression critical stress is nil or negligible compared to the diagonal tension stress, the panel reacts in pure diagonal tension (TDP).



σTDP

τ

α

TDP

TDP

σTDP =

τ = nominal shear stress

2τ sin 2α

The result is that the skin pulls on its edges, in both x and y directions thus inducing compression stresses in the members.

σTDP

σyp τ

TDP =

σxp

σx =

TDP

τ ;σy = τ tan α tan α

(these values are easily found using the Mohr circle) The normal forces induced in the members are: tB τAe tA τBe N TD ==cotan α ; NTD =tan α = tan α x y 2 2 2 2 tan α The angle α is such that the variation in internal energy of the system is minimal entailing the following result: ε − εx tan2 α = TDP ε TDP − ε y In which εTDP is the diagonal strain of the skin (tension) and εx and εy the horizontal and vertical members strains (compression).


Static stressing manual V1-1 • STIFFENED PANELS The incomplete diagonal tension theory was developed by Kuhn, Peterson and Levin for the NACA(1): Very often, the critical diagonal compression stress to which the skin shear critical stress corresponds is not negligible. The skin then reacts in stable shear for part of the load and in diagonal tension for the remaining part of the load:

τ

σt

σTD

TDI

TD

α σTD =

+

2 kτ sin 2α

Π/4

CS

σc σc = - σt = - (1 - k) τ

k is the "diagonal tension factor": k ∈ [0;1] k = 0: CS k = 1: TDP The stresses in the rectangular sheet result from overlaying CS + TD on a 2D-element of the base (x, y) are:

σyp τ σxp

1

TDI

kτ tan α σy = kτtan α

σx =

: NACA technical note 2661: "A summary of diagonal tension" Part 1 - Methods of analysis - May 1952.


Static stressing manual V1-1 • STIFFENED PANELS And on one 2D-element according to the theoretical folds:

σ1

TDI

σ2

τ12

2 kτ + (1 - k) τ sin 2α sin 2α σ2 = - (1 - k) τ sin 2α τ12 = (1 - k) τ cos 2α

σ1 =

In practice, α is close to 45°. Under these conditions, σ1 and σ2 are practically the principal stresses, giving:

τmax =

σ1 − σ 2 =τ 2

The incomplete diagonal tension theory is qualified as " theoretical engineering" by its authors because it is developed on the basis of theoretical analyses (in particular those of Wagner) and readjusted following a great number of tests. However from these tests it appears that the skin continues to be pure shear loaded beyond τcp. Therefore, the diagonal tension factor k is calculated using an empirical formula. Furthermore, if the members are fastened to the skin, a portion of the skin must be subjected to the same compression stresses as these members. The load-carrying width calculation formula is obtained from this, which is also empirical, and is given in this method. These load-carrying widths must not, in any case, be confused with those from Karman (compressed panels). In TDP, they tend to zero, the entirety behaving as a frame on which a film has been stretched. In TDI, the stresses of the centre plane of skin bays are not constant especially if these bays are in an end span (for example, at the edge of an opening). In this latter-mentioned case, the edge members bend and the diagonal tension tends to concentrate in the tensioned diagonals.


Static stressing manual V1-1 • STIFFENED PANELS For these reasons, the term "incomplete" diagonal tension is used instead of semi-diagonal tension considered as the overlaying of a pure shear state and a diagonal tension state (not pure). The fineness of the results is based on the two empirical formulas that have just been mentioned. Considering this, it is easy to understand why the use of this method is limited to the use domain specified in the following paragraph.

V1-1 •5.2-LIMITS The member flange on the skin side must not be too thin: es ≥ 0.6 e The proportions of the bays must be within the range: B A 0.2 ≤ ≤ 1 ; 120≤ ≤ 1500 A e Riveting: p 2.5 ≤ ≤ 10 D Pay attention to very thick skin bays (outside the domain): possibility of non-conservative forecasts. Stiffeners are approximately evenly spaced and the thicknesses of adjacent bays are similar.


Static stressing manual V1-1 • STIFFENED PANELS V1-1 •5.3-PRINCIPLE OF THE STIFFENED PANEL STRESSING UNDER INCOMPLETE DIAGONAL TENSION The theory of the incomplete diagonal tension was developed to calculate thin web beams consisting of flanges with the same section connected by a constant thickness web, which is stiffened by identical vertical members:

T

The shear load is constant along the beam and therefore the web bays, all identical, are subjected to the same constant shear flow. Under these conditions, the fold angle is the same in all bays. As seen in paragraph 5.1, α is such that the internal energy variation is minimum. To determine α, it is necessary to know the strain of the web in the fold direction, the strain of the flanges and the strain of the uprights. However, these values themselves depend on α. In practice, α is close to 45°. To solve this problem, an initial set of strain values is calculated assuming that the folds are at 45°, then the resulting tanα value is calculated. This calculation is then reiterated using the angle α calculated in the previous step until convergence. In general, three iterations suffice. In a general case, the shear flow varies from one point to another of the panel. In principle, it is known at the centre of each bay. Also, the thickness of the skin may vary from one bay to another as well as the sections of the members. However, there is no simple analytical method to calculate the variations of the tension field. Here it is necessary to use some simplifications making it possible to return locally to the simple configuration described above. This results in the calculation point notion. The information below makes it possible to estimate the static failure loads and stresses in stiffeners and in the skin at stiffeners. The section area of transverse members is only used to determine the fold angle. In and around the studied super-stiffener, the diagonal tension field is assumed uniform and the skin thickness, the section and the pitch distance of members constant.



Calculation point

y x

α

To determine the same effects at the transverse members, it may be necessary to repeat the same type of calculation considering the transverse members as stiffeners and vice-versa. The choice will be made considering the dimensional variations and the variations of the nominal shear field in the skin and the margins obtained. If the safety margins are low (less than 15%) it is necessary to confirm then by structural tests. This being said, and as we are concerned by what happens at a stiffener, we shall use the super-stiffener notion as in the compressed panel cases (refer to paragraphs 1.1, 2.2, 4.1 and 4.2). If the super-stiffener is not symmetrical in relation to the plane (G, x, z) it is necessary to build an equivalent symmetrical super-stiffener. Lt1.0

Lt2.0

Gp

Y d e1

G Gr

e2 dr

z


Y

dp

Static stressing manual V1-1 • STIFFENED PANELS The index 0 concerns the initial section in the following equations: e=

e Lt + e 2 Lt 2.0 e1 + e 2 ; Lt0 = 1 1.0 e1 + e 2 2

Diagonal tension factor: k =

æ1 + kö Loading rate: Rs = ç ÷ è1 − kø e τ + e2 τ2 τ= 1 1 e1 + e 2

k1 + k 2 2

2.3

V1-1 •5.4-LOADING RATE, DIAGONAL TENSION FACTOR • The loading rate of a mesh Rs is the ratio between the nominal shear stress and the critical buckling stress. • The diagonal tension factor formula is as follows: æ ö τ çç R s = < 1÷÷ Þ (k = 0) τ cp è ø

æ R s 0. 4343 − 1ö (Rs ≥ 1) Þ ç k = 0.4343 ÷ Rs + 1ø è

V1-1 •5.5-LOAD-CARRYING SECTIONS AND WIDTHS TD

Lt

TD

Lt

σxp σp

e

Lt0

Lt0



Index 0 relates to the initial section, i.e. before folding.

• Load-carrying widths (NACA formula): LtTD = (1 - k)

Lt 0 2

Stiffener direction: Lt0 =

b a ; Transverse member direction: Lt 0= 2 2

• Load-carrying section: Exactly the same method is used as for the compressed super-stiffeners, i.e. the subtractive method: instead of rebuilding the load-carrying section from load-carrying widths, remove the initial section which is not load-carrying. ∆STD = 2 (Lt0 - LtTD) e = (1 + k) Lt0e STD = S0 - ∆STD The calculation of the moment of inertia of the load-carrying section is only useful in the stiffener direction.

INTEGRAL STIFFENER: e W TD = æç d 0 − ö÷ ∆STD è 2ø TD W dTD = d0 + TD S 2 æ e e2 ö ∆ITD = ç æç d 0 − ö÷ + ÷ ∆STD 2ø 12 ø èè

ITD = I0 - ∆ITD - (d0 - dTD)2 STD


Static stressing manual V1-1 • STIFFENED PANELS FASTENED STIFFENER:

W

I

TD YYp

TD YYp

e = WYYp0 + æç e t − ö÷ ∆STD è 2ø

2 ææ eö e 2 ö TD = IYYp0 - ç ç e t − ÷ + ÷ ∆S 2ø 12 ø èè

Then apply the rules of the corrected section property calculation given in paragraph 2.2 (different materials) assuming that materials are of the linear elastic type. æ S TD S p ö E = ç TD ÷ Ep + æç TDr ö÷ E r èS ø èS ø æ E p ö TD æ Er ö W TD ÷ W YYp + çè ÷ø W YYr YY = ç è Eø E TD W YY d = TD S æ E p ö TD æ Er ö I TD ÷ I YYp + çè ÷ø I YYr YY = ç è Eø E

TD

2

TD TD ITD = ITD d YY - S

V1-1 •5.6-STRESSES IN THE SUPER-STIFFENER V1-1 •5.6.1-General method The load, the average stress and the average compression strain induced by the diagonal tension in the super-stiffener is formulated as follows: N TD =x

2 kLt 0 eτ N TD σ TD TD x x ;σ TD = ; ε = x x S TD E tan α x x

In fact, compression is not constant along the stiffener. It is at the minimum level at the transverse members (gusset effect) and in the middle, at maximum: σ TD x min σ TD x

σ TD bö x min æ = (1 - k) ç1.78 − 0.64 ÷ + k ; TD ≥ 1 è aø σx


Static stressing manual V1-1 • STIFFENED PANELS The same applies in the transverse member direction:

N

TD y

= - kaeτ tan α ;σ

σ TD y min σ TD y

TD y

=

N TD y S TD y

;ε

TD y

=

σ TD y Ey

σ TD aö y min æ = (1 - k) ç1.78 − 0.64 ÷ + k ; TD ≥ 1 è bø σy

The diagonal tension strain in the skin is: æ 2k ö τ εTD = ç + (1 − k ) (1 + ν) sin 2 α÷ è sin 2 α ø Ep The fold angle in relation to the stiffeners is such that: tan2 α =

ε TD − ε TD x ε TD − ε TD y

V1-1 •5.6.2-Simplified approaches As the fold angle is close to 45°, it is assumed that sin2α is equal to 1. This gives:

εTD = ((1 + k) +ν (1 - k))

τ Ep

ASSUMPTION 1: If the stiffening ratio and the stiffener material are similar to that of the transverse members, the folds are considered to be at 45° whatever k is. ASSUMPTION 2: If the transverse members can be considered as being infinitely stiff, TD determine σ x using the chart on the following page.

Remark: This assumption is not conservative for the stiffener as the folds axis tends to turn to the most rigid members direction.



CHART GIVING

σ TD x , WHEN: τ

• The transverse members are infinitely rigid. • ν = 0.33 S Plot k on the X axis, and æç x0 − 1ö÷ on the Y axis. If necessary, interpolate to obtain the è be ø required value.

0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6

1

0.7

σ TD x τ

0.8

0.9

0.9

1

0.8

1.1

(S0/be) - 1

0.7

1.2

0.6 1.4 0.5

1.6 1.8 2

0.4 0.3

2.5 3 4 5

0.2 0.1 0 0

0.1

0.2

0.3

0.4

0.5 k

0.6

0.7

0.8

0.9


1

Static stressing manual V1-1 • STIFFENED PANELS V1-1 •5.7-STRESSES IN THE SKIN V1-1 •5.7.1-In the pockets The pockets assumed identical on either side of the stiffener are subjected to in plane stress: kτ.tan α τ TDP

kτ/tan α y x

Tresca criterion: 2

τ max æ k ö ∏ö æ æ τ max ö = 1+ç = 1÷ ÷ ;ç α = ÷ Þ ç è τ ø è τ 4ø è tan 2α ø

V1-1 •5.7.2-At the stiffener In the middle between the transverse members, the maximum compression in stiffener direction is: σ TD xp min =

Ep E

σ TD x min

The skin is also highly stressed in the cut-out plane passing through the skin-stiffener fastener. Here, the Tresca criterion in the hatched section is: τ'max e = 1.3 τ et

æ k ö 1+ç ÷ è1 + kø

2

et e


Static stressing manual V1-1 • STIFFENED PANELS V1-1 •5.8-STRESSES IN THE STIFFENER Compression in the stiffener is at the maximum in the middle between the transverse members and is: E r TD σ TD σ x min r min = E Also, the average compression stress of the stiffener is: E σ TD = r σ TD r x E As indicated in paragraph 1.4, these calculations are only applicable to typical areas. There are no stiffener run-outs and stiffeners react in single compression. The case of stiffener run-outs is discussed in Chapter V1-4 ("THIN WEB BEAMS").

V1-1 •5.9-ALLOWABLE STRESSES AT UL V1-1 •5.9.1-Local buckling The inter-rivet buckling failure mode does not exist when the skin reacts in diagonal tension. A forced crippling mode substitutes this mode which is described in the next paragraph. Therefore, natural local buckling can only occur in the stiffener (refer to Chapter V1-3).

V1-1 •5.9.2-Forced crippling In reality, this is amplified bending: because the stiffener skin side flange is loaded transversely by skin undulations, its free edge tends to follow these undulations whereas the edge it forms with the web remains straight. In this case, local bending stresses due to the transverse loads induced by the buckle pattern of the skin, amplified by compression are added to the compression stresses due to diagonal tension. With skin shear failure, forced crippling is one of the two main causes of failure. Refer to Chapter V1-3 ("STIFFENERS") to determine the allowable compression stressσflf and the corresponding nominal shear stress τflf.


Static stressing manual V1-1 • STIFFENED PANELS V1-1 •5.9.3-Super-stiffener column buckling • Strength with regard to column buckling is checked, in principle, by comparing the average compression stress σ TD and the critical column buckling stress. To calculate this, it is x necessary to take the specific features of the flat panels in diagonal tension into account. When stiffener bending starts, the diagonal tension attempts to oppose it. Reactions are as if the column was fastened over its entire length to an elastic foundation. The end fixity coefficient calculation given below takes this phenomenon into account: TD = LTD = KTD A ; K

1 Lt ö æ 1 + k2 ç3 − 4 0 ÷ è A ø

TD TD Then calculate σ TD , I , LTD) with the information given in paragraph V1-1•4. xcrit using (S

• Depending on the slenderness ratio of the super-stiffeners, it is possible that the column buckles under a two half wave mode. For this reason, it is advisable to check that σ TD x min does

LTD not exceed the buckling stress calculated with the slenderness ratio . 2ρ TD

V1-1 •5.9.4-Skin failure Generally, the skin is thicker at stiffeners than at the bottom of bays. Failure may then occur either at the bay-pad blend radii or in the stiffener-skin fastener rows. • Failure in the pocket (TRESCA):

τmax allowable =

σR 2

• Failure at skin-stiffener fasteners:

τ' max allowable =

σR 2

V1-1 •5.10-SKIN STIFFNESS As long as the skin remains stable, the skin shear stiffness is represented by Coulomb’s modulus, G. In diagonal tension, it is appreciably more elastic. The effective transverse shear modulus is GTDI. As we conceive incomplete diagonal tension as overlaying of a single shear state and of a diagonal tension state, the following can be formulated: æ 1 1− k k ö (γTDI = γCS + γTD) ⇔ ç = + ÷ G G TD ø è G TDI Flat stiffened panels under shear - Page V1-1•5/16 revision 2 © AEROSPATIALE 1998

Static stressing manual V1-1 • STIFFENED PANELS with: tan 2 α E 4 2 cot 2 α = + + S0 S mt 1 − k sin 2 2α G TD − (1 − k ) + Lt 0 e Ae 2

Giving: TD ö ε TD æ 4ν ö G TDI æ x cot α + ε y tan α ÷ = çç1 + ç 2 − 1÷ k + νE p ÷ τ G è sin 2α ø è ø

−1

SPECIAL CASES:

æ ç ç G • (α = 45°) Þ ç G TDI = σ TD p ç 1 + ( 4 ν − 1) k + 2 ν ç τ è

ö ÷ ÷ ÷ ÷ ÷ ø

• Infinitely stiff transverse members: G GTDI ≈ E p σ TD x 1 + ( 4 ν − 1) k + ν cot α E τ

with,

σ τ

TD x

obtained from the graph and:

σ τ

TD x

S0 − (1 + k ) Lt 0 e 2k

V1-1 •5.11-GENERAL INSTABILITY, DESIGN These panels are protected from the general buckling risk for as long as the stiffeners are designed to rupture under forced crippling or under super-stiffener column buckling under a nominal shear load slightly less than the shear strength of the skin. Inversely, if the static margin depends on the skin shear strength, then our panel is protected from any general instability risk. This condition is satisfied if the stiffeners have sufficient inertia and super flange thickness. If, in addition, the skin comprises pads of a sufficient thickness, then the static margin is only governed by the shear strength of the skin material: τallowable =

σR 2

The study condition is satisfied if:

e ≤ 0.7 et


Static stressing manual V1-1 • STIFFENED PANELS V1-1 •5.12-EXAMPLE Use the example in paragraph 2.3 (also see paragraphs 4.5.5 and 4.6.4). The nominal shear at UL in the bays adjacent to the stiffener is: τ = 160 MPa The properties of the skin (2024PLT3) are: Ep = 70300 MPa ;σRp = 440 MPa ;σ0.2p = 270 MPa The properties of the stiffeners and transverse members (7075PLT6) are: Er = 73800 MPa ;σRr = 495 MPa ;σ0.2r = 420 MPa The two materials have the same Poisson ratio: ν = 0.33

SECTION IN STIFFENER DIRECTION: 170 =

= 3

2 35 =

=

25 3 30

2 4 16

Stiffener:

Skin:

Sr = 185 mm²

Sxp0 = 375 mm²

IYYr = 62186 mm4

IYYp0 = 1485 mm4

dr = 13.9 mm

dp0 = 1.86 mm

WYYr = 2572 mm3

WYYp0 = - 698 mm3


Static stressing manual V1-1 • STIFFENED PANELS SECTION IN TRANSVERSE MEMBER DIRECTION:

Small section: Smt = 509 mm² ; y0 S = 1604 mm² ; yp0 S = 1095 mm² 24 4 24 4

90

2.5 120

8 3 3

6 2

35 530

The small section is the one where the stiffeners pass. Between the stiffeners, the section of the transverse member alone is 509 mm². This value is used to calculate the fold angle and the stresses in the transverse member halfway from the stiffeners, giving: S = 1604 mm² Smt = 509 mm² ; y0

POCKET FOLDS

Refer to the example in paragraph 3.3 (a1 = a2 = A = 530 mm ;1 b= b2 = B = 170 mm): τcp1 = τcp2 = 52 MPa LIMITS, AS PER PARAGRAPH 5.2

Stiffener flange: es 3 = > 0.6 e 2


Static stressing manual V1-1 • STIFFENED PANELS Skin-stiffener attachment with 4.8 diameter rivets, pitch distance 25 mm: p 2.5 < æç = 5.2ö÷ < 10 èD ø Skin bay proportions: B æ A = 265ö < 1500 0.2 < æç = 0.32ö÷ < 1 ; 120 < ç ÷ èA ø èe ø

LOADING RATE, DIAGONAL TENSION FACTOR (paragraph 5.4) 160 τ = = 3.077 52 τ cp

k=

3.077 0. 4343 − 1 = 0.24 3.077 0. 4343 + 1

LOAD-CARRYING WIDTHS AND SECTIONS AT UL (paragraph 5.5)

• Stiffener direction: ∆S TD = (1 + k) Lt0e = 1.24 x 85 x 2 = 211 mm² x TD S x = S0 - ∆STD = 560 - 211 = 349 mm² Sr = 185 mm² ; TD S = 164 mm² xp W

eö æ = W TD = - 276 mm3 ÷ ∆S TD YYp0 + ç e t = x è ø 2 2 ææ eö e2 ö 4 TD = ITD + e − + ç ÷ ç ÷ ∆S x = 572 mm YYp0 t 2ø 12 ø èè

TD YYp

I TD YYp

164 ö æ 185 ö 73800 = 72154 MPa Ex = æç ÷ 70300 + ç ÷ è 349 ø è 349 ø æ 70300 ö 276 + æ 73800 ö 2572 = 2361 mm3 W TD ÷ ç ÷ YY = - ç è 72154 ø è 72154 ø 2361 = 6.76 dTD = 349 æ 70300 ö 572 + æ 73800 ö 62186 = 64161 mm4 I TD ÷ ç ÷ YY = ç è 72154 ø è 72154 ø ITD = 64161 - 349 x 6.76² ≈ 48200 mm4


Static stressing manual V1-1 • STIFFENED PANELS • Transverse member direction: ae = (1 + k) = 1.24 x 530 = 657 mm² ∆S TD y 2 TD TD S y = S0 - ∆S y = 1604 - 657 = 947 mm² Smt = 509 mm² ; TD S = 438 mm² yp 438 ö æ 509 ö 73800 = 72181 MPa Ey = æç ÷ 70300 + ç ÷ è 947 ø è 947 ø

STRESSES IN THE SUPER-STIFFENER, TD ANGLE AT UL (paragraph 5.6)

• Compression in stiffener direction: 2 kLt 0 eτ N TD == 2 x 0.24 x 85 x 2 x 160 x cot α = - 13019 cot α x tan α N TD σ TD TD TD x σ x = TD = - 37.3 cot α ;ε x = x = - 0.052%.cot α Sx Ex • Compression in transverse members: N TD = - kaeτ tan α = - 0.24 x 530 x 2 x 160 x tan α = - 40590 tan α y σ

TD y

=

N TD y S TD y

= - 42.9 tan α ;ε

TD y

=

σ TD y Ey

= - 0.059%.tan α

• Diagonal tension (refer to paragraph 5.6.2): 160 = 0.339% εTD ≈ (1 + 0.24 + 0.33 (1 - 0.4)) 70300

• Diagonal tension angle: æ ε = ε TD 0.339 + 0.52 ö r (α0 = 45°) → ç tan 2 α 1 = TD = ÷ TD ε TD − ε mt 0.339 + 0.059 ø è æ 0.339 + 0.052 / 0.9903 ö (α1 = 44.72°) → ç tan 2 α 2 = ÷ 0.339 + 0.059 x 0.9903 ø è æ 0.339 + 0.052 / 0.9916 ö (α2 = 44.76°) → ç tan 2 α 3 = ÷ 0.339 + 0.0591 x 0.9916 ø è

α = 44.7° ; tanα = 0.99


Static stressing manual V1-1 • STIFFENED PANELS • Super-stiffener stresses: σ TD =x

37.3 = - 37.6 MPa 0.99

σ TD b x min = (1 - k) æç1.78 − 0.64 ö÷ + k = 0.76 (1.78 - 0.64 x 0.32) + 0.24 = 1.44 TD è aø σx σ TD x min = - 54.1 MPa REMARK 1: The stiffening ratios in the stiffener and transverse member direction are similar and the panel is a little stiffer in the stiffener direction.

560 − 170 x 2 1604 − 530 x 2 = 0.39 ; = 0.34 560 1604 The use of the following approximation (deviation negligible in relation to the general solution) is justified: = - 37.3 MPa ;σ TD (α ≈ 45°) Þ (σ TD x x min = - 1.44 x 37.3 = - 53.7 MPa) REMARK 2: Use of the assumption "infinitely rigid transverse members" and therefore the chart in paragraph 5.6.1 gives the following results (- 6% deviation): æ æ S0 ö ö ö æ σ TD çç . ÷ → ç x = 0 22 . ; σ TD = − 0 22. x 160 = − 35 2 .MPa÷ ÷ − 1 = 0.65 ;k = 0 24 x ø è τ è è 2 Lt 0 e ø ø

(- 37.3 cot α = - 35.2) Þ (α = 46.7°) POCKET STRESSES (refer to paragraph 5.7.1) (α ≈ 45°) Þ (τmax = τ = 160 MPa) SKIN STRESS AT THE RIVET ROW (refer to paragraph 5.7.2)

2 0.24 ö τ' 1 + æç ÷ max = 1.3 è 1.24 ø 3

2

x 160 = 0.883 x 160 = 141.2 MPa <τmax

Therefore, skin failure will occur at the pocket blend radius. SKIN COMPRESSION IN STIFFENER DIRECTION E p TD 70300 54.1 = - 52.7 MPa σ TD σ x min = xp min = 72155 Ex


Static stressing manual V1-1 • STIFFENED PANELS STRESSES IN THE STIFFENER (refer to paragraph 5.8) E r TD 73800 σx = − 37.6 = - 38.5 MPa 72155 Ex E 73800 = r σ TD 54.1 = - 55.3 MPa x max = − 72155 Ex

σ TD = r σ TD r min

ALLOWABLE STRESSES AT UL

• Local stiffener buckling (refer to paragraph 4.5.5): flange: 504 MPa, web: 476 MPa • Forced crippling (refer to example in paragraph V1-3 5.4): σflf = 144 MPa ;τflf = 283 MPa • Super-stiffener column buckling (refer to paragraph 5.9.3): 1 1 KTD = = = 0.94 Lt 0 ö 85 ö 2 æ 2 æ 1 + 0.24 ç 3 − 4 1 + K ç3− 4 ÷ ÷ è è 530 ø A ø LTD = 0.94 x 530 = 498 mm Johnson formula (refer to paragraph 4.6.4): S TD TD x = LTD TD = 42.4 σ0 = 328 MPa ;λ CE I 328 2 x 42.4 2 TD λ =λ CE :σ TD = 328 = 260 MPa xcrit 4 ∏ 2 72155 λTD λ = CE : σ TD xcrit = 311 MPa 2 These values calculated with the slenderness ratio at UL are conservative as λ is decreasingly monotonic with the applied load. Therefore, with regards to stability, the allowable is given by the forced crippling.

• Skin failure at the pocket (refer to paragraph 5.9.4): σ Rp 440 = τmax allowable = = 220 MPa 2 2


Static stressing manual V1-1 • STIFFENED PANELS STATIC MARGIN:

τ max allowable

=

220 = 1.38 160

τ max τ flf 283 = = 1.77 160 τ

In our example, the static margin results from the comparison of nominal shear stress at UL with the failure stress of the skin material. The other margins are not very different, meaning that the skin pad and the super flange of the stiffener are correctly sized. The risk of general buckling of the panel is nil and the weight is close to the optimum.


Static stressing manual V1-1 • STIFFENED PANELS V1-1 •6.9-STRESSES IN FRAMES The single maximum compression stress in frames riveted to the skin mid-way between stringers is: σ TD mt min =

E mt TD σ ymin E TD y

If frames are floating, apply the bending moments given in paragraph 6.6. The loads then apply to the cross sections of frames alone.

V1-1 •6.10-ALLOWABLE STRESSES AT UL V1-1 •6.10.1-Local buckling Natural local buckling can only occur in stiffeners and in frames (refer to Chapter V1-3).

V1-1 •6.10.2-Forced crippling Proceed in the same manner as for flat panels (refer to paragraph 5.9.2). Refer to Chapter V1-3 "STIFFENERS" to calculate the allowable stresses. Check the strength of frame and stringer outer flanges.

V1-1 •6.10.3-Super-stiffener column buckling Check the strength with regard to column buckling by comparing the maximum compression stress σ TD x min and the critical buckling stress, calculated assuming that the super-stiffener is clamped at frames. TD TD Calculate σ TD xcrit using (S x ; I ) (refer to the Johnson formula in paragraph V1-1 •4).

V1-1 •6.10.4-Skin failure Generally, the skin is thicker at stiffeners than in pockets bottoms. Failure may then occur either in the skin pad-pocket blend radii connection fillets or in the skin-stiffener fastener rows. Curved stiffened panels in shear - Page V1-1•6/14 revision 2 © AEROSPATIALE 1998

Static stressing manual V1-1 • STIFFENED PANELS • Pocket failure (TRESCA):

τmax allowable =

σR (0.65 + ∆) 2

• Failure at skin-stiffener fasteners: Ditto for τmax allowable With: æS ö æS ö ∆ = 0.3 tanh ç mt ÷ + 0.1 tanh ç r ÷ è Ae ø è Be ø The skin pads have to be included in member cross sections.

V1-1 •6.11-SKIN STIFFNESS Skin shear stiffness, for as long as the skin is stable, is represented by Coulomb’s modulus, G. It is appreciably more compliant in diagonal tension. The effective transverse modulus of elasticity is GTDI: ε TD cot α + ε TD tan α æ 4ν ö G x y =1+ ç 2 − 1÷ k + νE p G TDI τ è sin 2α ø

V1-1 •6.12-GENERAL INSTABILITY, FRAME LOCAL BUCKLING, DESIGN The nominal shear stress causing cylinder failure by general buckling τIG is obtained using the chart on the following page. Frames must be designed so that the static margin of the panel is given by another criterion. General instability must be impossible. Furthermore, frame flanges shall be of a sufficient thickness so that the allowable load is not limited by local buckling or forced crippling. In fact, it could be potentially dangerous in one way or another for the general static margin to be the same as the margin of a frame (frames).

Curved stiffened panels in shear - Page V1-1•6/15 revision 2 © AEROSPATIALE 1998

Static stressing manual V1-1 • STIFFENED PANELS USE OF THE CHART: B • æç < 1ö÷ : Calculate ρx and ρy, assuming that the entire sheet interacts with the stringers and èA ø the frames (even floating). B • æç > 1ö÷ : Calculate ρx and ρy, assuming that the sheet does not interact with the stringers èA ø and the frames (member sections only). If ρx > 8.6 mm, take ρx = 8.6 mm.

The solid straight lines correspond to the domains covered by the tests (good correlation). No explanation has been found for the abrupt transition from one to the other.

Do not extrapolate outside the scope of the chart.

Use the Young's modulus in compression.

General instability criterion for curved panels in shear

τ IG x 103 E

3 2.5 2 1.5 1 0.5 0 0

2

4

6

(

ρx ρy

r

3 4

)

8

10

12

7 8

x 104

AB


Static stressing manual V1-1 • STIFFENED PANELS V1-1 •6.13-EXAMPLE Use again the example in paragraph 5 with r=2820 mm: Nominal shear at UL in skin bays adjacent to the stiffener is: τ = 160 MPa The properties of the skin (2024PLT3) are: Ep = 70300 MPa ;σRp = 440 MPa ;σ0.2p = 270 MPa ;ν = 0.33 The properties of the stringers and the frames (7075PLT6) are: Er = 73800 MPa ;σRr = 495 MPa ;σ0.2r = 420 MPa ;ν = 0.33

ASSOCIATED STRINGER AND SKIN:

170 =

= 3

2 35 =

Stiffener: Sr = 185 mm² IYYr = 62186 mm4 dr = 13.9 mm WYYr = 2572 mm3

= Skin: Sxp0 = 375 mm² IYYp0 = 1485 mm4 dp0 = - 1.86 mm WYYp0 = - 698 mm3

25 3 30

2 4 16


Static stressing manual V1-1 • STIFFENED PANELS ASSOCIATED FRAME AND SKIN:

24 4 24

4

90

2.5 120

8 3 3

6 2

35 530

The small section of the frame is the one at the stringer passage. Between the stringer passages, the section area of the frame itself is 509 mm². This value is used to calculate the angle of the folds and the stresses in the frame at mid-distance from the stringers, giving: S = 1604 mm² Smt = 509 mm² ; y0 Properties of the small section:

Frame: Sc = 369 mm² Ic = 355232 mm4 dc = 84.4 mm

Frame + skin Sy0 = 1464 mm² I0 = 2415091 mm4 d0 = 19.9 mm


Static stressing manual V1-1 • STIFFENED PANELS POCKET FOLDING

Refer to example in paragraph 3.3 (a1 = a2 = A = 530 mm ;1 b= b2 = B = 170 mm ; R = 2820 mm): τcp1 = τcp2 = 57 MPa LIMITS AS PER PARAGRAPH 6.2

Stiffener inner flange: es 3 = > 0.6 2 e Skin-stiffener connected by 4.8 diameter rivets, pitch distance 25 mm: p 2.5 < æç = 5.2ö÷ < 10 èD ø Skin bay proportions (A>B): B æ A = 265ö < 1500 ; A <2820 0.2 < æç = 0.32ö÷ < 1 ; 120 < ç ÷ èA ø èe ø 3 LOADING RATE, DIAGONAL TENSION FACTOR (paragraph 6.4) æ 0.2 < B < 0.5ö Þ æ ϕ = 0.4343 + 521.2 x 2 ö ç ÷ ç ÷ è ø è A 2820 ø ϕ = 0.8039 160 τ Rs = = = 2.807 57 τ cp

2.807 0.8039 − 1 k= = 0.393 2.807 0.8039 + 1

For the identical flat panel subjected to the same nominal shear stress (refer to the example in paragraph 5), the diagonal tension factor is 0.24. Here, the effect of the diagonal tension is more significant even though the folds occur a little later. LOAD-CARRYING SECTION AND WIDTHS AT UL (paragraph 6.5)

• Stringer direction: ∆S TD = (1 + k)Lt0e = 1.393 x 85 x 2 ≈ 237 mm² x TD S x = Sx0 - ∆S TD = 560 - 237 = 323 mm² x S = 138 mm² Sr = 185 mm² ; TD xp


Static stressing manual V1-1 • STIFFENED PANELS W

eö æ = W TD = - 224 mm3 ÷ ∆S TD YYp0 + ç e t − x è 2ø 2 ææ eö e2 ö TD = IYYp0 - ç ç e t − ÷ + ÷ ∆S TD = 459 mm4 x è ø 2 12 ø è

TD YYp

I TD YYp

138 ö æ 185 ö 73800 = 72303 MPa E TD = æç ÷ 70300 + ç ÷ x è 323 ø è 323 ø æ 70300 ö 224 + æ 73800 ö 2572 = 2407 mm3 W TD ÷ ç ÷ YY = - ç è 72300 ø è 72300 ø 2407 dTD = = 7.45 323 æ 70300 ö 459 + æ 73800 ö 62186 = 63920 mm4 I TD ÷ ç ÷ YY = ç è 72300 ø è 72300 ø ITD = 63925 - 323 x 7.45² ≈ 46000 mm4

• Frame direction: The frames are fastened to the skin, giving:

530 x 2 = 738 mm² 2 = 1604 - 738 = 866 mm²

∆S TD = 1.393 x y S TD y

509 ö æ 509 ö x 73800 = 72357 MPa = æç1 − E TD ÷ x 70300 + ç ÷ y è ø è 866 ø 866

STRESS AND STRAIN, STRINGER AND FRAME DIRECTION, TD ANGLE AT UL (paragraph 6.6)

• Compression, stiffener direction: kbeτ = - 0.393 x 170 x 2 x 160 x cot α = - 21357 cot α tan α N TD σ TD x x = TD = - 66.1 cot α ;ε TD = = - 0.091%.cot α x Sx E TD x

N TD =x σ TD x


Static stressing manual V1-1 • STIFFENED PANELS • Compression, transverse member direction: = - 0.393 x 530 x 2 x 160 x tan α = - 66584 tan α N TD y σ TD = y

N TD y S TD y

= - 76.9 tan α ;ε TD = y

σ TD y E TD y

= - 0.106%.tan α

• Diagonal tension: æ 2 x 0.393 ö 160 εTD = ç + (1 + 0.33) (1 − 0.393) sin 2 α÷ è sin 2 α ø 70300 0.179% εTD = + 0.184% x sin 2 α sin 2 α

• Diagonal tension angle: æ ö ç ÷ 2 TD TD ε − εx æç B < 1ö÷ Þ ç tan 2 α = ÷ ; 1 æç B ö÷ = 0 015% . 2 èA ø ç 1 æ B ö ÷ 24 è R ø TD ε TD − ε y + ç ÷ ÷ ç è 24 è R ø ø Considering the small amount of flattening strain, α0 = 45° was selected: æ ö 0.363 + 0.091 (α0 = 45°) → ç tan 2 α 1 = ÷ 0.363 + 0.106 + 0.015 ø è → (α1 = 44.1° ;α2 = 44.3° ;α3 = 44.2°)

α = 44.2° . tan α = 0.974 • Super-stiffener stresses: 66.1 = - 67.8 MPa 0.974 b = (1 - k) æç1.78 − 0.64 ö÷ + k = 0.61 x (1.78 - 0.64 x 0.32) + 0.39 = 1.35 è aø

σ TD =x σ TD x min σ TD x

σ TD x min = - 91.5 MPa


Static stressing manual V1-1 • STIFFENED PANELS • Super-frame stresses: σ TD = - 76.9 x 0.974 = - 74.9 MPa y æ σ TD ö aö æ ö y min æ ç (1 − k ) çè1.78 − 0.64 ÷ø + k = 0.26 < 1÷ Þ çç TD = 1÷÷ è ø b è σy ø

σ TD y min = - 74.9 MPa

• Bending moments in the super-stiffener: MTD = 0.393 x

170 x 530 2 x 160 x 2 x 0.974 = 86325 N.mm 24 x 2820

SKIN POCKET STRESSES (refer to paragraph 6.7.1) (α ≈ 45°) Þ (τmax = τ = 160 MPa) SKIN STRESS AT THE ROW OF RIVETS (refer to paragraph 6.7.2) 2

2 0.393ö τ' 1 + æç ÷ x 160 = 0.9 x 160 = 144 MPa <τmax max = 1.3 x è 1.393 ø 3 Skin failure will occur therefore in the pocket blend radius.

SKIN COMPRESSION, STIFFENER DIRECTION, BETWEEN FRAMES E p æ TD M TD TD ö σ TD = σ − (d + e t )÷ p TD TD ç x max ø Ex è I 70300 æ 86325 σ TD x ç − 91.5 − x ( 7.45 + 3)ö÷ = - 108 MPa p = ø 72303 è 46000


Static stressing manual V1-1 • STIFFENED PANELS STRESSES IN THE STIFFENER (refer to paragraph 6.8)

• At mid-distance from frames: E r TD 73800 σx = − x 67.8 = - 69.2 MPa TD 72303 Ex E r TD 73800 σ TD σ x max = − x 91.5 = - 93.4 MPa r min = TD 72303 Ex Outer flange: E r æ TD 73800 æ 86325 M σ TD − TD d TD ö÷ = x 7.45ö÷ = - 107.7 MPa ç − 91.5 − çσ rs = TD è x max ø 72303 è ø 46000 Ex I Inner flange: 73800 æ 86325 σ TD x (30 − 7.45)ö÷ = - 50.2 MPa ç − 91.5 + rt = ø 72303 è 46000

σ TD = r

• Inner flange stress at frames: 73800 æ 86325 σ TD (30 − 7.45)ö÷ = - 112.5 MPa ç − 67.8 − rt = ø 72303 è 46000

STRESSES IN FRAMES (refer to paragraph 6.9)

σ TD mt min = - 74.9 x

73800 = - 76.4 MPa 72357

ALLOWABLE STRESSES AT UL (paragraph 6.10)

• Local stiffener buckling (refer to paragraph 4.5.5): Inner flange: 504 MPa, web: 476 MPa • Stringer forced crippling (refer to example in paragraph V1-3•5.4): σflf = - 177 MPa ;τflf = 213 MPa • Frame forced crippling (refer to paragraph 6.10.2): Discussed in example V1-3•5.4 σflf = - 171 MPa ;τflf = 258 MPa No danger from this standpoint.


Static stressing manual V1-1 • STIFFENED PANELS • Super-stiffener column buckling (refer to paragraph 6.10.3): KTD = 0.5 LTD = 0.5 x 530 = 265 mm Johnson formula (refer to paragraph 4.6.4): S TD σ0 = 328 MPa ;λ = L = 22.2 I TD æ TD ö 328 2 x 22.2 2 TD λ = λ → σ = − = 309 MPa÷ 328 ( ç crit CE ) 2 4 ∏ 72303 è ø These values calculated with slenderness at UL are conservative as λ is monotonic, decreasing with the applied load. TD CE

TD

With regards to stability, the allowable value is given by the forced crippling.

• Skin pocket failure (refer to paragraph 6.10.4): σ Rp τmax allowable = (0.65 + ∆) 2 æ 544 ö æ 220 ö ∆ = 0.3 tanh ç ÷ + 0.1 tanh ç ÷ = 0.2 è 530 x 2 ø è 170 x 2 ø τmax allowable =

440 x 0.85 = 187 MPa 2

STATIC MARGIN: τ max allowable 187 = = 1.17 160 τ τ flf 213 = = 1.33 160 τ

FRAME SIZING (paragraph 6.12)

ρx =

I0 = Sx 0

ρy =

2415091 = 40.6 mm (small frame section) 1464

58759 = 10.2 mm (refer to paragraph 2.3) 560

7 æ ö ç ρx ρy 8 ÷ æτ ö 4 x 10 = 16.8÷ Þ ç IG x 10 3 > 3÷ ç 3 è ø E ç r 4 AB ÷ è ø

(

)



To be conservative, put down E=Ep, giving: τIG > 211 MPa >τmax allowable

Therefore, the frames are sized in such a manner that general instability is not possible, which was what was required.






V1-1 •6-STRESSING OF CURVED STIFFENED PANELS IN SINGLE NOMINAL SHEAR

Ty

G z

x

REMINDER: The curvature of the panel, radius r, is in the plane normal to the stiffeners (refer to paragraph 1.4). This paragraph is dedicated to incomplete diagonal tension. For more information, refer to paragraph 5 (Calculation of flat stiffened panels in single nominal shear). The approach given here is based on theoretical analysis, readjusted following tests. It applies to the domain defined in the "Limit" paragraph.


Static stressing manual V1-1 • STIFFENED PANELS V1-1 •6.1-DIAGONAL TENSION THEORIES The analysis of the diagonal tension in curved panels is based on the methods developed for flat panels. Therefore, the goal here is to describe the problems posed by the panel curvature. The typical example of a cylindrical fuselage section subject to a torsion moment is a good basis for discussion. • Pure diagonal tension theory: If the cross section of the fuselage section was polygonal instead of circular, the diagonal tension theory in flat webs would be applicable with some minor modifications. This will also be the case if the circular cross section skin was supported by polygonal shaped frames as the diagonal tension flattens the skin bays. However, in the present case, the frames, i.e. the transverse members, are circular. Therefore, the diagonal tension in a bay no longer resides in one plane. To simplify the problem, Wagner established fundamental relations (equations) derived from the flat panel theory for two cases: a) Bays long in the axial direction, stringers (stiffeners) spaced close together: r A > 2B ; B < 3 b)Bays long in the transverse direction, frames (transverse members) spaced close together: r B > 2A ; A < 3

r B A

In the first case (a), the major portion of the diagonal tension is in the planes of a polygonal cylinder. In the second case (b), it is in hyperboloids of rotation. In the case of flat panels subjected to TDP (refer to paragraph 5.1), the angleα is such that the variation of internal energy of the system is minimal, giving the following result: tan2 α =

ε TDP − ε TDP x ε TDP − ε TDP y


Static stressing manual V1-1 • STIFFENED PANELS This principle can be applied to curved panels considering that: Case a: Flattening is unrestricted between two frames. The change from the circular cross section shape to the polygonal cross section is equivalent to a decrease in the spacing of stringers, ∆, equal to the difference in length between arc B and chord B : 1 æ B3 ö ∆≈ ç ÷. 24 è r 2 ø The angle of the folds is then given by: ε TDP − ε xTDP tan2 α = 2 1 æ Bö TDP ε TDP − ε y + ç ÷ 24 è r ø ∆ Remark: Strain is fictitious. It does not correspond to a stress in the frames. B Case b: In the case of long skin bays in the frame direction, the same type of reasoning results in: ε TDP − ε xTDP tan2 α = 2 1 æ Aö TDP ε TDP − ε y + ç ÷ tan 2 α 8è rø In practice, the first equation is used when A is greater than B and the second when B is greater than A. If the bays are "square bays", the same results should be obtained using the two equations, which implies: æ tan 2 α = 1 ö Þ (α = 30°) ç ÷ è 3ø The latter-mentioned value is quite representative. In practice, the fold angle is from 20° to 45°. Another difference in behaviour between flat and curved panels due to skin bays flattening: the "roof" effect at the stringers implies that the stringers are subject to radial loads. The diagonal tension bends the curved panel stringers whereas it acts against stiffener bending in flat panels.


Static stressing manual V1-1 • STIFFENED PANELS Assuming that these radial loads are steadily distributed, there is: B

B

q

r

A

q

B BA 2 BA 2 q =τe tan α : M(0) = M(A) = τe tan α ;M æ A ö = τe tan α ç ÷ r 12 r 24 r è 2ø With the sign convention: (M > 0: compressed skin). However, if the frames are fastened to the skin, there is no roof effect around the frames as they remain curved. And if the bending moment mid-way between the frames is effectively equal to M(A/2), its modulus is much greater than M (A) at the frames. • Incomplete diagonal tension theory1: Diagonal tension develops earlier in curved panels than in flat panels. The diagonal tension factor k is given by generalising the experimental formula applicable to flat panels. In stringer direction, the calculation formula of the load-carrying width is identical to the one applicable to flat panels. The same applies to the deformation equationεTD. There are two types of frames: frames directly connected to the skin and frames that are not connected to the skin, which are named "floating frames". Normally, frames are connected to the skin and slotted where stringers pass. In this case, the centripetal load applied to them due to the diagonal tension is uniform. They work in the vault manner, meaning that they are subjected to compression stresses only. The flanges of these frames are likely to be destroyed under forced crippling. Here, the load-carrying section is calculated like in the stringer direction. Floating frames are only loaded at frame-stiffener intersections. Owing to this, they are subjected to bending moments. Furthermore, no load-carrying skin reacts and folds are free to develop on either side of the frames. 1

: NACA technical note 2661: "A summary of diagonal tension" Part 1 - Methods of analysis - May 1952


Static stressing manual V1-1 • STIFFENED PANELS The use of floating frames can only be considered for panels with bays that are long in the stringer direction (a) type). If the frames are fastened to the skin, the bending moment of super-stringers at frames is considered as being equal to that mid-way from frames, from a modulus standpoint. If the frames are floating, this bending moment value is doubled.

V1-1 •6.2-LIMITS This method can be used in the design of panels made of aluminium alloys normally used in the aeronautical industry. Use the rules applicable to flat panels (paragraph 5) if the curvature radius is greater than 3800 mm. These calculations only apply to typical areas (no stiffener run-outs) as indicated in paragraph 1.4. The member flange (skin side) must not be too thin: es ≥ 0.6 e The bays proportions shall be in the following range: r æ B < 1ö → æ 0.2 ≤ B ≤ 1 ;120 ≤ A ≤ 1500 ; A < ö÷ ç ÷ ç èA ø è 3ø A e r æ B > 1ö → æ 0.2 ≤ A ≤ 1 ;120 ≤ B ≤ 1500 ; B < ö÷ ç ÷ ç èA ø è 3ø B e Riveting: 2.5 ≤

p ≤ 10 D

Pay attention to very thick skin bays (outside field of application): possible non-conservative estimates. The stiffeners are practically evenly spaced and the thicknesses of adjacent bays similar.


Static stressing manual V1-1 • STIFFENED PANELS V1-1 •6.3-CALCULATION PRINCIPLE FOR STRESSING STIFFENED PANELS IN INCOMPLETE DIAGONAL TENSION To determine α, the strain of the web according to the fold axis, the strain of the stringers and the strain of the frames needs to be known. However, these values themselves depend on α. In practice, α is between 20 and 45°. To solve this problem, an initial set of strain values is calculated assuming fold angle to be α = α0, then the resulting value of tanα is calculated. This calculation is then reiterated using angle α calculated at the previous pitch up to convergence. Three iterations suffice if α0 is properly selected. There is no simple analytical method to calculate the variations in the tension field as indicated for flat panels. Here it is necessary to use a simple approach making it possible to compare the present problem to a simple configuration. The information given below makes it possible to estimate the static failure loads and stresses in stringers and in the skin at stringers. The frame sections were, in particular, used to determine the fold angle. In and around the studied super-stiffener, the diagonal tension field is assumed to be uniform. The thickness and the curvature radius of the skin, the section and the spacing of members is assumed to be constant.

Calculation point

y x α

Calculations cannot be repeated considering the frames as stiffeners and the stringers as transverse members due to the curvature. Experimental data on frame failure under diagonal tension is not available. Frames will always be sized so that failure occurs either in the skin or in the stringers. Frames are considered to be correctly sized when their local buckling does not affect the general margin and overall instability of the fuselage. If safety margins are low (less than 15%) it is mandatory to confirm them by structural tests.


Static stressing manual V1-1 • STIFFENED PANELS The super-stiffener notion will be used as for the flat panels (refer to paragraph 5.3). The curvature of the skin is assumed to be sufficiently small for it not to be taken into account when calculating the section properties. An equivalent symmetrical super-stiffener has to be built if the super-stiffener is not symmetrical in relation to the plane (G,x,z): Lt1.0

Lt2.0

Gp

Y d

G

e1

Gr

Y

dp dr

e2

h

z In the following equations, the index 0 relates to the initial section. e Lt + e 2 Lt 2.0 e1 + e 2 ; Lt0 = 1 1.0 e1 + e 2 2 r + r2 r= 1 2

e=

Diagonal tension factor: k =

k1 + k 2 2

1

æ1 + kö ϕ Loading rate: Rs = ç ÷ è1 − kø

τ=

e1 τ1 + e 2 τ 2 e1 + e 2

V1-1 •6.4-LOADING RATE, DIAGONAL TENSION FACTOR • The loading rate of a skin bay Rs is the ratio between the nominal shear stress and the critical buckling stress.


Static stressing manual V1-1 • STIFFENED PANELS • The general diagonal tension factor k is formulated as follows: æ ö τ çç R s = < 1÷÷ Þ (k = 0) τ cp è ø

æ R sϕ − 1ö (Rs ≥ 1) Þ ç k = ϕ ÷ R s + 1ø è with: æ 0.5 < B < 1ö Þ æ ϕ = 0.4343 + 260.6 ç ÷ ç è ø è A æ 0.5 < A < 1ö Þ æ ϕ = 0.4343 + 260.6 ç ÷ ç è ø è B

eA ö ÷ rB ø eB ö ÷ rA ø

æ 0.2 < B < 0.5ö ü ç ÷ï è øï æ A ý Þ çè ϕ = 0.4343 + 521.2 æ 0.2 < A < 0.5ö ï ç ÷ è ø ïþ B

eö ÷ rø

V1-1 •6.5-LOAD-CARRYING SECTIONS AND WIDTHS V1-1 •6.5.1-Stiffener direction Lt

Lt

σxp e σp Lt0

Lt0

• Load-carrying widths (NACA formula): LtTD = (1 - k)

Lt 0 2


Static stressing manual V1-1 • STIFFENED PANELS • Load-carrying section: A procedure identical to the one use for compressed super-stiffeners is used, i.e. the subtractive method. Rather than rebuilding the load-carrying section using load-carrying widths, remove the initial section which is not load-carrying. ∆S TD = 2 (Lt0 - LtTD) e = (1 + k) Lt0e x S TD = Sx0 - ∆S TD x x INTEGRAL STIFFENER: e W TD = æç d 0 − ö÷ ∆S TD x è 2ø W TD dTD = d0 + TD Sx 2 æ e e2 ö ∆ITD = ç æç d 0 − ö÷ + ÷ ∆S TD x 2ø 12 ø èè

ITD = I0 - ∆ITD - (d0 - dTD)2 S TD x

FASTENED STIFFENER:

W

e = WYYp0 + æç e t − ö÷ ∆S TD x è 2ø 2 ææ eö e2 ö = IYYp0 - ç ç e t − ÷ + ÷ ∆S TD x 2ø 12 ø èè

TD YYp

I TD YYp

Then apply the design rules for the properties of the corrected section given in paragraph 2.2 (different materials) assuming that materials are of the linear elastic type. E

TD x

W

æ S TD æ S ö xp ö = ç TD ÷ E p + ç TDr ÷ E r è Sx ø è Sx ø

TD YY

æ E p ö TD æ E r ö = ç TD ÷ W YYp + ç TD ÷ W YYr è Ex ø è Ex ø

dTD =

TD W YY S TD x

æ E p ö TD æ E ö I I TD + ç TDr ÷ IYYr YY = ç TD ÷ YYp è Ex ø è Ex ø 2

TD ITD = ITD dTD YY - S x


Static stressing manual V1-1 • STIFFENED PANELS V1-1 •6.5.2-Frame direction Proceed in the same manner if frames are fastened to the skin: ∆S TD = (1 + k) y

ae 2

S TD = Sy0 - ∆S TD y y E

TD y

æ S TD æ S mt ö yp ö ÷÷ E mt = çç TD ÷÷ E p + çç TD è Sy ø è Sy ø

Calculation of the moment of inertia of the load-carrying section is not necessary. It is even simpler if frames are floating: = Smt S TD y E TD = Emt y

V1-1 •6.6-SUPER-MEMBER STRESSES The compression force, average stress and average strain in the stringer direction induced by diagonal tension is formulated as follows: N TD =x

N TD σ TD kbeτ TD x x = ; ε = ;σ TD x x TD tan α E S TD x x

Compression is not constant over the entire length of a stringer. Compression is minimum at the frames (gusset effect) and maximum mid-way between frames: σ TD x min σ TD x

σ TD bö x min æ = (1 - k) ç1.78 − 0.64 ÷ + k ; TD ≥ 1 è aø σx

The same is calculated in frame direction:

N

TD y


σ TD y min σ TD y

TD y

=

N TD y S TD y

;ε

TD y

=

σ TD y

E TD y

σ TD aö y min æ = (1 - k) ç1.78 − 0.64 ÷ + k ; TD ≥ 1 è ø σy b Curved stiffened panels in shear - Page V1-1•6/10 revision 2

© AEROSPATIALE 1998

Static stressing manual V1-1 • STIFFENED PANELS Diagonal tension strain in the skin is: æ 2k ö τ εTD = ç + (1 − k )(1 + ν)sin 2α÷ è sin 2α ø Ep

The fold angle in relation to stiffeners is such that: æ ç ε TD − ε xTD æç B < 1ö÷ Þ ç tan 2 α = èA ø ç 1 ε TD − ε TD + ç y è 24

ö ÷ ÷ 2 B æç ö÷ ÷ è r ø ÷ø

æ ö ç ÷ TD ε TD − ε x æç B > 1ö÷ Þ ç tan 2 α = ÷ 2 èA ø ç ÷ 1 A ε TD − ε TD + æç ö÷ tan 2 α ÷ ç y è ø 8è rø æ A = 1ö Þ æ tan 2 α = 1 ;α = 30°ö ç ÷ ç ÷ èB ø è ø 3

• Bending moment in the super-stiffener half way from frames: M æTDA ö = k ç ÷ è 2ø

BA 2 τe tan α 24 r

This moment compresses the stringer skin side flange and the skin. • Bending moments in the super-stiffener at frames: • Frames fastened to the skin:

M (TD A) = - k


• Floating frames:

M (TD A) = - k


This moment compresses the stringer inner flange. • Bending moments in floating frames mid-way from stringers: M æTDB ö = - k ç ÷ è 2ø

B2 A τe tan α 24 r

This moment compresses the frame inner flange. Curved stiffened panels in shear - Page V1-1•6/11 revision 2 © AEROSPATIALE 1998

Static stressing manual V1-1 • STIFFENED PANELS • Bending moments in floating frames at stringers: M (TD B) = k


This moment compresses the skin.

V1-1 •6.7-STRESSES IN THE SKIN V1-1 •6.7.1-In pockets kτ.tan α τ

TDP

kτ/tan α y x

Tresca criterion:

τ max æ k ö = 1+ç ÷ τ è tan 2 α ø

2

V1-1 •6.7.2-At the stringer In the middle between transverse members, the compression stress is: σ

TD xp

=σ

σ TD xp =

TD x min

-

M (TD A / 2) I

TD

dTD, if integral stiffener

E p æ TD M (TD ö A/2) ç σ x min − TD (d TD + e t )÷ , if fastened stiffener TD Ex è I ø

The skin is also highly stressed in the cutting plane crossing the skin-stiffener fastener rows. Here, the Tresca criterion in the hatched part is: τ max e = 1.3 τ et

æ k ö 1+ç ÷ è1 + kø

2



e et

V1-1 •6.8-STRESSES IN THE STIFFENER • In the middle between two transverse members, the maximum single compression stress, the outer flange stress (fastened stiffeners) and the inner flange stress are respectively equivalent to: σ TD r min =

Er σ TD x min TD Ex

σ TD rs =

M (TD ö E r æ TD A/2) d TD ÷ σ − ç x min TD TD Ex è I ø

σ TD rt =

M (TD ö E r æ TD A/2) h − d TD )÷ σ + ç ( x min TD TD Ex è I ø

• The average compression stress of the stiffener is: σ TD = r

Er σ TD x E TD x

• At frames, the outer flange stress is: σ TD rt =

TD ö E r æ TD M ( A ) h − d TD ) )÷ σ + ç x TD TD ( Ex è I ø

Reminder: If frames are floating, the bending moment is twice as great as the bending moment at mid-distance from the frames.



V1-1 •7-STRESSING OF STIFFENED PANELS UNDER COMPRESSION AND SHEAR

Ty

G z

P = - Nx x

REMINDER: The curvature of the skin is in the plane normal to the stiffeners (refer to paragraph 1.4). To calculate the skin bay folding load and the failure load of a fastened skin-stiffener panel under combined compression and shear it is first of all necessary to determine these same values in pure compression (refer to paragraph 4) then in single nominal shear (refer to paragraph 5 or 6). The following is then obtained: • skin buckling stresses (σcp0 ;τcp0), • failure loads (Pallowable,0 ;τflf ;τRp). It is assumed that the skin failure load in single nominal shear τRp is not modified significantly by the presence of a compression load. On the other hand, strain induced by diagonal tension folds in fastened stiffeners outer flange affects the cross sections of the superstiffeners and therefore the allowable compression load. In other words, forced crippling interacts with column buckling. Naturally, this type of failure does not occur in integral stiffener panels. Here it is accepted that the initial instability of the stiffener (local buckling or column buckling) encountered when the panel is subjected to single shear interacts with the column buckling.

Stiffened panels under compression and shear - Page V1-1•7/1 revision 2 © AEROSPATIALE 1998

Static stressing manual V1-1 • STIFFENED PANELS V1-1 •7.1-STRESSING PRINCIPLES The method developed here is from Kuhn, Peterson and Levin1. • Under combined loading, the folds in the skin bays occur earlier. The buckling stresses (σcp;τcp) are obtained using the interaction formula: σ cp σ cp 0

2

æ τ cp ö σ σ ÷÷ = 1, and before folding: cp = p + çç τ cp τ è τ cp 0 ø

• If the failure mode of the super-stiffener in single nominal shear is forced crippling of the stiffener outer flange, then the failure load is obtained by the interaction formula: æτ ö Pallowable + ç allowable ÷ Pallowable .0 è τ flf ø

1. 5

P P = 1 ; and: allowable = τ allowable τ

If the failure mode of the super-stiffener under single nominal shear is skin failure, then the interaction curve above is truncated by the vertical line of abscissa τRp: P Pallowable.0

interaction curve

Pallowable P allowable loads

τ

τallowable

τRp

0

1

τ

τflf

: NACA technical note 2661: "A summary of diagonal tension" Part 1 - Methods of analysis - May 1952


Static stressing manual V1-1 • STIFFENED PANELS Integrally stiffened panels: The single shear load causing local buckling or column buckling of the super-stiffener has to be determined. This is an iterative calculation. Then it is possible to proceed as above replacing τflf by τbuckling. In all cases, the ideal situation is to determine τflf (or τbuckling) by testing. • The calculation of stresses in the stiffeners and the transverse members under a given load is more complex. The premature occurrence of folds affects the diagonal tension factor k. The presence of a compression stress increases the angle of folds (90° in single compression, 20 to 45° in single shear, depending on the panel curvature). Furthermore, the load-carrying section under compression is not equal to the effective section under shear. On the other hand, it is accepted that there is no difference in nature between compression stress due to P (direct) and the stress due to diagonal tension (indirect). The total stress is simply the addition of the two. If the super-stiffener is not symmetrical in relation to the plane (G,x,z) it is necessary to build an equivalent symmetrical super-stiffener:

Lt1,0

Lt2,0

Y

d

e1

Gp G

h

Y

dp e2

Gr

dr

z In the following equations, the index 0 relates to the initial section. e=

e Lt = e 2 Lt 2.0 e1 + e 2 ; Lt0 = 1 1.0 e1 + e 2 2

σ cp1.0 + σ cp 2.0 ö æ 1 1æ 1 1 öö æ = ç + ÷÷ ç σ cp 0 = ÷ Þç è ø 2 è R c 2 è R c1 R c 2 ø ø e τ + e2 τ2 τ= l 1 2e

Diagonal tension factors in single shear: k0 =

k 1.0 + k 2.0 2

1

æ1 + k0 ö ϕ Rs = ç ÷ è1 − k0 ø


Static stressing manual V1-1 • STIFFENED PANELS V1-1 •7.2-LIMITS Those given in paragraphs 4, 5 and 6.

V1-1 •7.3-LOADING RATE, DIAGONAL TENSION FACTOR V1-1 •7.3.1-Loading rate • The loading rate in single compression is: P Rc = Pcp0 Remark: A load ratio and not a stress ratio has to be used to eliminate non-linearity between applied load and stresses. Pcp0 is the beginning of fold load. • The loading rate in single nominal shear is: τ Rs = τ cp0 • The loading rate under combined loads is: R= τ cp τ cp 0

R c + R 2c + 4 R s2 τ P = = 2 Pcp τ cp

=

Pcp R s σ cp R ; = c = R σ cp 0 R Pcp 0

V1-1 •7.3.2-Diagonal tension factor The general diagonal tension factor is formulated as below:


Static stressing manual V1-1 • STIFFENED PANELS (R < 1) Þ (k = 0) æ R ϕ − 1ö (R ≥ 1) Þ ç k = ϕ ÷ R + 1ø è with: æ 0.5 < B < 1ö Þ æ ϕ = 0.4343 + 260.6 eA ö ç ÷ ç ÷ è ø è A rB ø æ 0.5 < A < 1ö Þ æ ϕ = 0.4343 + 260.6 eB ö ç ÷ ç ÷ è è B ø rA ø æ 0.2 < B < 0.5ö ü ç ÷ï e è øï æ A Þ ç ϕ = 0.4343 + 521.2 ö÷ ý æ 0.2 < A < 0.5ö ï è rø ç ÷ è ø ïþ B

V1-1 •7.4-LOAD-CARRYING CROSS SECTIONS AND WIDTHS V1-1 •7.4.1-Stiffener direction Under the combined load (P, τ), the load-carrying widths are: • Compression: LtC = Lt0 • Shear: LtTD =

σ cp σp

,

Rs Lt (1 − k ) 0 R 2

And variations to the corresponding sections are: • Compression: ∆S Cx = 2 (Lt0 - LtC)e = 2 (Lt0 - LtTD)e • Shear: ∆S TD x The properties of the load-carrying sections are then determined as previously in compression or single shear (refer to paragraphs 2.2, 4.2 and 5.5). If the stiffener and the skin are made of different materials, the corrected section design rules given in paragraph 2.2 are used. Under compression, the secant modulus of the equivalent homogeneous material is obtained using the secant modulus of the materials determined for the compression mode alone. Under shear, use the same skin and stiffener secant moduli in the interpolation law. æ S Cxp ö æ S TD æ Sr ö æ S ö xp ö C TD E sx = ç C ÷ Esp + ç C ÷ Esr ; Esx = ç TD ÷ Esp + ç TDr ÷ Esr è Sx ø è Sx ø è Sx ø è Sx ø Stiffened panels under compression and shear - Page V1-1•7/5 revision 2 © AEROSPATIALE 1998

Static stressing manual V1-1 • STIFFENED PANELS V1-1 •7.4.2-Frame direction If the frames are fastened to the skin: ∆S TD = (1 + k) y

ae 2

S TD = Sy0 - ∆S TD y y E

TD y

æ S TD æ S mt ö yp ö ÷÷ E mt = çç TD ÷÷ E p + çç TD è Sy ø è Sy ø

If the frames are floating: S TD = Smt y E TD = Emt y

V1-1 •7.5-STRESSES IN SUPER-MEMBERS The compression load, average stress and strain, in the stringer direction, induced by diagonal tension, are formulated as follows: N TD =x

N TD σ TD kbeτ TD x x ;σ TD = ; ε = x x TD TD S E tan α x sx

The stress and strain, in the stringer direction, induced by the single compression load, are formulated as follows: N Cx σ Cx C σ = C ;ε x = C Sx E sx C x

The force, the average stress and the strain, in the stringer direction, are formulated as follows: + NCx ;σx = σ TD + σ Cx ;εx = ε TD + ε Cx Nx = NTD x x x The compression due to τ is not constant along a stringer. It is at a minimum level at the frames (gusset effect) and maximum mid-way from the frames: σ TD σ TD bö x min x min æ = (1 - k) ç1.78 − 0.64 ÷ + k ; TD ≥ 1 ;σx min = σ Cx + σ TD x min TD è ø a σx σx Frame direction: N

TD y


TD y

=

N TD y S TD y

;ε

TD y

=

σ TD y E TD y


Static stressing manual V1-1 • STIFFENED PANELS σ TD y min σ TD y

σ TD aö y min æ = (1 - k) ç1.78 − 0.64 ÷ + k ; TD ≥ 1 è σy bø

The diagonal tension strain in the skin is: æ 2k ö τ εTD = ç + (1 − k ) (1 + ν) sin 2 α÷ è sin 2 α ø E sp The fold angle in relation to the stiffeners is such that: 2 æ 1 R s æ Bö ö ε TD − ε y + ç ÷ ÷ ç 24 R è r ø ÷ æç B < 1ö÷ Þ ç cot 2 α = èA ø ç ÷ ε TD − ε x ç ÷ è ø 2 æ ö 1 Rs æ Aö ε TD − ε y + ç ÷ tan 2 α ÷ ç 8 R è rø æç B > 1ö÷ Þ ç cot 2 α = ÷ èA ø ç ÷ ε TD − ε x ç ÷ è ø

• Bending moment in the super-stiffener, mid-way from the frames: M æTDA ö = k ç ÷ è 2ø


This moment compresses the stringer outer flange and the skin. • Bending moments in the super-stiffener, at the frames: • Frames fastened to the skin:

M (TD A) = − k


• Floating frames:

M (TD A) = − k


This moment compresses the stringer inner flange. • Bending moments in floating frames, mid-way from stringers: M æTDB ö = − k ç ÷ è 2ø


This moment compresses the frame inner flange. Stiffened panels under compression and shear - Page V1-1•7/7 revision 2 © AEROSPATIALE 1998

Static stressing manual V1-1 • STIFFENED PANELS • Bending moments in floating frames, at stringers: M (TD B) = k


This moment compresses the skin. Riveted frames: no bending moment in frames.

V1-1 •7.6-STRESSES IN THE SUPER-STIFFENER V1-1 •7.6.1-In the skin In the middle between transverse members, the compression in stringer direction is: σxp min = σx min -

M (TD A/2) I

TD

d TD , if integral stiffener

E sp æ M (TD ö A/ 2) σxp min = C ç σ x min − TD ( d TD + e t ) ÷ , if fastened stiffener E sx è I ø

V1-1 •7.6.2-In the stiffener • The average compression stress of the stiffener is: E σr = Csr (σx) E sx • In the middle between transverse members, the maximum compression stress, the outer flange stress (fastened stiffeners) and the inner flange stress are respectively: E σr min = Csr σx min E sx σrs =

M (TD ö E sr æ A/2) d TD ÷ σ − ç x min C TD E sx è I ø

σ TD rt =

M (TD ö E sr æ A/2) + h − d TD )÷ σ ç ( x min C TD E sx è I ø

• At frames, the flange stress is: σrt =

M (TD ö E sr æ A) h − d TD )÷ σ + ç x C TD ( E sx è I ø


Static stressing manual V1-1 • STIFFENED PANELS Reminder: If frames are of the floating type, the bending moment is twice as high as the moment at mid-distance from frames.

V1-1 •7.7-STRESSES IN FRAMES In frames fastened to the skin, the maximum compression at mid-distance from frames is: E mt TD σmt min = TD σ y min Ey

If frames are floating, apply the bending moments given in paragraph 7.5. The loads are then applied to the frame cross section alone.

V1-1 •7.8-ALLOWABLE LOADS V1-1 •7.8.1-At limit load The folding load/ratio limit load ratio is: Pcp τ cp P = = PLL τ LL PLL R

V1-1 •7.8.2-At ultimate load Generally, a shear stress (abscissa)-compression load (ordinate) diagram is used. The combined failure load (τallowable, Pallowable) is, as indicated in paragraph 7.1, at the intersection of the interaction curve and the straight line passing through points (0,0) and (τ,P). æτ ö Pallowable + ç allowable ÷ Pallowable .0 è τ crit .0 ø

1. 5

P P = 1 ; maxτ{ {Rp;τcrit.0}; = allowable allowable} = min τ τ τ allowable

The same result is obtained when the following dimensionless diagram is used: æ τ ö P Mark the point ç ; ÷ , add, if necessary, the vertical straight line of the abscissa è τ crit .0 Pallowable .0 ø τ Rp . The allowable load is determined as below. τ crit.0

FASTENED STIFFENERS:

τcrit.0 = τflf

INTEGRAL STIFFENERS:

τcrit.0 = τbuckling



Compression-shear (failure) interaction diagram 1 0.9 0.8 0.7 P P0

0.6 0.5 0.4 0.3 0.2 0.1 0

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

T/T0


1

Static stressing manual V1-1 • STIFFENED PANELS V1-1 •7.9-EXAMPLE Use the example in paragraph 4.6.4 and 5.12 (flat panel: r → ∞). The load applied to the super-stiffener at UL is: Nx = - 38000N ;τ = 160 MPa The properties of the skin (2024PLT3) are: Ep = 70300 MPa ;σRp = 440 MPa ;σ0.2p = 270 MPa The properties of the stiffeners and transverse members (7075PLT6) are: Er = 73800 MPa ;σRp = 495 MPa ;σ0.2r = 420 MPa The Poisson ratio of the two materials is the same: ν = 0.33 CROSS SECTION, STIFFENER DIRECTION: 170 =

= 3

2 35 =

=

25 3 30

2 4 16

Stiffener:

Skin:

Sr = 185 mm²

Sxp0 = 375 mm²

IYYr = 62186 mm4

IYYp0 = 1485 mm4

dr = 13.9 mm

dp0 = - 1.86 mm

WYYr = 2572 mm3

WYYp0 = - 698 mm3


Static stressing manual V1-1 • STIFFENED PANELS CROSS SECTION, TRANSVERSE MEMBER DIRECTION:

Small section: Smt = 509 mm² ; y0 S = 1604 mm² ; yp0 S = 1095 mm² 24 4 24 4

90

2.5 120

8 3 3

6 2

35 530

The small section is the one at the stiffener passage. Between the passages, the cross section area of the transverse member alone is 509 mm². This value is used to calculate the angle of the folds and the stresses in the transverse member at mid-distance from stiffeners, giving: S = 1604 mm² Smt = 509 mm² ; y0


Static stressing manual V1-1 • STIFFENED PANELS FAILURE LOAD, MARGIN AT UL (paragraph 7.8.2)

• Column buckling load in single compression:

Pcrit.0 = 90200 N

• Forced crippling load (single shear):

τflf = 283 MPa

• Skin bay failure shear:

τRp = 220 MPa

• Compression-shear interaction: 1.5 Pallowable æ τ allowable ö Pallowable 38000 = +ç = 237.5 ÷ = 1 ; maxτ{ allowable} = 220 MPa ; 90200 è 283 ø τ allowable 160 Use of the dimensionless diagram: ì τ ü 220 160 τ P 38000 = = 0.565 ; max = 0.777 ; = = 0.421 í ý= τ flf 283 Pcrit.0 90200 î τ flf þ 283 MS =

OM - 1 = 14% Om

Compression-shear (failure) interaction diagram 1 0.9 0.8

P P0

0.7 0.6

M

0.5 0.4 0.3 0.2 0.1

m

0 0

0.2

0.4

0.6

0.8

T/T0


Static stressing manual V1-1 • STIFFENED PANELS FOLDING OF POCKETS UNDER SINGLE LOADS

τcp0 = 52 MPa ;σcp0 = 36 MPa ; cp0 P = 20492 N LOADING RATE, DIAGONAL TENSION FACTOR (paragraph 7.3)

Rs =

160 38000 = 3.077 ; cR= = 1.854 52 20492

1.854 2 + 1.854 2 + 4 x 3.077 2 = 4.141 2 ϕ = 0.4343 4.1410. 4343 − 1 = 0.299 k= 4.1410. 4343 + 1 R=

POCKET FOLDING æ R s 3.077 ö = = 0.743÷ Þ (τcp = 0.743 x 52 = 38.6 MPa) ç è R ø 4.141 æ R C 1.854 ö σcp = 16.1 MPa) = = 0.448÷ Þ (Pcp = 0.448 x 20492 = 9175 N ; ç è R ø 4.141 æ P = 38000 = 25330 Nö Þ folding at: 9175 = 36% LL ç LL ÷ è ø 25330 1.5

LOAD-CARRYING SECTIONS AND WIDTHS AT UL (paragraph 7.4)

• Stiffener direction: Modelling of material curves and load-carrying sections in compression epsilonp sigmap epsilonr sigmar Epsilon Sigmap Sigmar Sigma 0.00023 16.1 0.00023 16.8 0.023% 16.1 16.9 16.4 36.0 37.8 36.7 0.00051 36 0.00051 37.8 0.051% 0.00149 104.73 0.00149 110 0.149% 104.7 110.1 107.6 epsilon Sp S P Esp Esr Es 0.023% 375.0 560.0 9164 70300 73800 71456 0.051% 262.4 447.4 16437 70300 73800 71747 0.149% 168.3 353.3 38001 70181 73800 72076 (These tables are compiled in the same way as those on page V1-1•4/19, with σcp = 16.1.)


Static stressing manual V1-1 • STIFFENED PANELS S Cx = 353 mm² Sr = 185 mm² ; TD S = 168 mm² xp C E sx = 72076 MPa Esp = 70181 MPa ; srE= 73800 MPa 85 LtTD = 0.743 x (1 - 0.299) x = 22,1 mm 2 = 2 (85 - 22.1) x 2 = 251 mm² ∆S TD x TD S x = 560 - 251 = 309 mm² Sr = 185 mm² ; TD S = 124 mm² xp

124 ö E sxTD = æç ÷ x 70181 x è 309 ø

æ 185 ö x 73800 = 72351 MPa ç ÷ è 309 ø

• Frame direction (fastened to the skin): 530 x 2 = 689 mm² 2 = 1604 - 688 = 915 mm²

∆S TD = 1.299 x y S TD y

509 ö æ 509 ö x 73800 = 72246 MPa = æç1 − E TD ÷ x 70300 + ç ÷ y è è 915 ø 915 ø

STRESSES AND STRAIN, STRINGER AND FRAME DIRECTION, TD ANGLE AT UL (paragraph 7.5)

• Compression, in stiffener direction, due to TD: kbeτ = - 0.299 x 170 x 2 x 160 x cot α = - 16266 cot α tan α − 16266 − 52.6 = = = - 52.6 cot α ;ε TD = - 0.073% cot α x 309 72351

N TD =x σ TD x

• Single compression load, in stiffener direction: N Cx = - 38000 N σ Cx = - 107.6 MPa ;ε Cx = - 0.149%


Static stressing manual V1-1 • STIFFENED PANELS • Compression, stiffener direction: Nx = - 16266 cot α - 38000 σx = - 52.6 cot α - 107.6 ;ε TD = - 0.073% cot α - 0.149% x • Compression, transverse member direction: N TD = - 0.299 x 530 x 2 x 160 x tan α = - 50730 tan α y σ

TD y

=

N TD y S TD y

= - 55.4 tan α ;ε

TD y

=

σ TD y E TD y

= - 0.077%.tan α

• Diagonal tension: æ 2 x 0.299 ö 160 εTD = ç + (1 + 0.33) (1 − 0.299) sin 2 α÷ è sin 2 α ø 70181 0.136% εTD = + 0.213% x sin2α sin 2 α

• Diagonal tension angle: æç B < 1ö÷ Þ æ cot 2 α = ε TD − ε y ö Þ α = 48.6° ; tanα = 1.136 ç ÷ èA ø ε TD − ε x ø è

• Super-stiffener stresses: 52.6 + σ Cx = - 46.4 - 107.6 = - 154 MPa = - 46.4 MPa ;σx = σ TD x 1136 . b = (1 - k) æç1.78 − 0.64 ö÷ + k = 0.7 x (1.78 - 0.64 x 0.32) + 0.3 = 1.4 è aø

σ TD =x σ TD x min σ TD x

C σx min = σ TD σ TD x min = - 65.1 MPa ; x min + σ x = - 172.7 MPa

• Transverse super-member stresses: σ TD = - 55.4 x 1.136 = - 62.9 MPa y æ σ TD ö aö æ ö y min æ ç (1 − k ) çè 1.78 − 0.64 ÷ø + k < 1÷ Þ çç TD = 1÷÷ è ø b è σy ø

σ TD y min = - 62.9 MPa


Static stressing manual V1-1 • STIFFENED PANELS SKIN COMPRESSION, STIFFENER DIRECTION (paragraph 7.6)

σp min =

C E sp

E

C s

σx min = - 172.7 x

70181 = - 168.3 MPa 72076

STRESSES IN THE STIFFENER

E srC 73800 σr min = C σx min = x 172.7 = - 176.9 MPa 72076 Es STRESSES IN THE FRAMES (refer to paragraph 7.7)

σ TD mt min =

E mt 73800 σy min = - 62.9 x = - 64.3 MPa TD 72246 Ey






YES

X

NO


1 4

5 4

2 1 2

Volume 1

5


Purpose

Scope


Contents

Document Manager



Not applicable.

V1 - 1 V1 - 2 V1 - 3 V1 - 4 V1 - 5 V1 - 6 V1 - 7 V1 - 8 V1 - 9



Validation

Name: J. HUET



3page 1

Title - Annex

Reference documents


Abbreviations




Definitions


Highlights

Issue

Date

Pages modified

A

02/98

V1 - 1 à V1 - 3 V1 - 7 à V1 - 9

B

05/99

V1 - 7


V1 - 4

New chapter.

V1 - 1


C

11/99




New chapter.

MTS 004 Iss. C

3Ann. page



List of approval

Dept. code

Function

BTE/CC/CM

Name / First name


Key words

Calcul

Bibliography

Néant

Signature

CAZET G.

Distribution list

Dept. code

Function


BQP/TE

Archives Diderot

SIBADE Alain

BQP/TE


SIBADE Alain


BOUTET Fernand

BTE/SM/MG



MTS 004 Iss. C

page IG1

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS CONTENTS revision Date V1-2 V1-2•1 V1-2•2 V1-2•3 V1-2•4

Buckling of thin plates and shells General information Flat plates Curved plates Cylinders and other shells

1 1 1 1 1

07/1996 07/1996 07/1996 10/1996 03/1997

Original issue Original issue Original issue Original issue Original issue

1- GENERAL 1.1- Introduction (V1-2•1/1)* 1.2- Single and combined loading (V1-2•1/2) 1.3- Plasticity corrections (V1-2•1/3) 1.4- Field of application (V1-2•1/4) 2- FLAT PLATES A - Summary of cases dealt with (V1-2•2/2) B - Flat plate analysis procedure (V1-2•2/4) 2.1- Single loading (V1-2•2/5) 2.2- Plasticity correction factors (V1-2•2/6) 2.3- Buckling factors (V1-2•2/7) 2.3.1- Rectangular flat plates 2.3.2- Flat parallelogram plates 2.3.3- Flat triangular plates 2.3.4- Flat plates with evolutive width

2.4- Combined loading (V1-2•2/27) 2.4.1- Combined loading without transverse loading 2.4.2- Combined loading with transverse loading

2.5- Examples (V1-2•2/38) 3- CURVED PLATES A - Summary of cases dealt with (V1-2•3/2) B - Curved plate analysis procedure (V1-2•3/3) 3.1- Single loading (V1-2•3/4) 3.2- Plasticity correction factors (V1-2•3/5) 3.3- Buckling factors (V1-2•3/6) 3.3.1- Rectangular curved plates

3.4- Combined loading (V1-2•3/13) 3.4.1-Shear + compression (or tension) 3.4.2- Compression + side pressure 3.4.3- Shear + side pressure

3.5- Examples (V1-2•3/15) *

: Page number between parentheses

Revision 1 © AEROSPATIALE - 1998

Contents

page V1-2.i

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS 4- CYLINDERS AND OTHER SHELLS A - Summary of cases dealt with (V1-2•4/2) B - Circular cylinder analysis procedure (V1-2•4/3) C - Elliptical cylinder analysis procedure (V1-2•4/4) D - Spherical shell analysis procedure (V1-2•4/5) E - Conical shell analysis procedure (V1-2•4/5) 4.1- Behaviour of circular cylinders under compression (V1-2•4/6) 4.1.1- Introduction 4.1.2- Linear theory and test results

4.2- Plasticity correction factors (V1-2•4/8) 4.3- Buckling factors (single loading) (V1-2•4/9) 4.3.1- Circular cylinders 4.3.2- Elliptical cylinders 4.3.3- Spherical shells 4.3.4- Conical shells

4.4- Interaction equations (combined loading) (V1-2•4/28) 4.4.1- Circular and elliptical cylinders 4.4.2- Conical shells

4.5- Examples (V1-2•4/33)


Contents

page V1-2.ii

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS SYMBOLS USED a: Bay length, long axis of an ellipse

kpl: Buckling factor of flat circular plates

b: Bay width, short axis of an ellipse

L: Length

Ci: Constants (curve equations)

m: number of half waves

-0.15

C= 0.6*In (3.748*(r/e)

) (constant)

MS: Static margin

Cf: 1.3*C (constant in cylinder bending)

Mf: Bending moment

UL: Ultimate loads

Mf,cr: Critical bending moment

LL: Limit loads

Mt: Torsion moment

d: Diameter

Mt,cr: Critical torsion moment

e: Plate or cylinder thickness

Mapp: Applied moment

E: Young's modulus

n: Stress-strain shape factor (Ramberg and

Es: Secant modulus

Osgood)

Et: Tangent modulus

n: Number of circumferential waves

G: Shear modulus

n0: Number of circumferential waves when

Gs: Secant shear modulus

L=r

I: Bending stiffness constant (inertia)

p: Pressure

J: Torsion stiffness constant

Pint: internal pressure

K,k: Buckling factors

Pcr: Critical pressure (external)

Kc,kc: Compression buckling factors

Pc,cr: Critical compression load

Ks,ks: Shear buckling factors

Papp: Applied load

Kf,kf: Bending buckling factors

r : Radius or small radius r: Critical curvature radius

Kt,kt: Torsion buckling factors

Ky,ky: Buckling factors under external radial Ri: Loading rate for load i alone pressure

Rc: Loading rate in single compression

Kp,kp: Buckling factors under hydrostatic Rs: Loading rate in single shear pressure Rf: Loading rate in single bending * * * k c ,k f ,k p : Buckling factors at the border of Rp: Loading rate under pressure two curves Rcs: Loading rate under a shear load kc0: Buckling factors of column plates under Rct: Loading rate in torsion single compression


Contents

page V1-2.iii

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS R: Large radius or loading rate under combined loads Sa: Strength modulus of the elliptic cylinder cross section Se: Strength modulus of the elliptic cylinder cross section r t= 1 − ν 2e (constant) e v: Distance to the neutral axis ycr: Critical curvature radius position from the neutral axis L2 Z= 1 − ν 2e (constant) re Z* = Z constant at the border of two curves αβγ: Exponents in interaction curve equations α: Also used to designate angles 4Z β: 2 (used here as a constant) π t ρ: Minimum curvature radius ρmoy: Average curvature radius ε: Strain (expansion) Φ: Flow η: Plasticity factor νe: Poisson's ratio - elastic νp: Poisson's ratio - plastic ν: Poisson's ratio - elastic - plastic σ: Normal stress σ0: Reference stress for transverse loading σa: Stress at position a σ0.2: Conventional allowable compressive yield stress σR: Tension failure stress σeq: Equivalent normal stress σcr: Critical stress


σiadm,0: Buckling stress corresponding to load i alone σcc,0: Critical single compression stress σcf,0: Critical single bending stress σcy,0: Critical stress under hydrostatic or radial pressure σcp,0: Critical stress under external pressure σcxe: Elastic longitudinal critical stress σcye: Elastic transverse critical stress σcx: Longitudinal critical stress corrected and taking transverse load into account σc,cr: Critical compression stress σf.cr: Contrainte critique en flexion σc,cr: Critical compression stress σc,cr: Critical compression stress σt: Longitudinal tension stress ∆σcc,0: Increase in critical compression stress due to internal pressure ∆σcf,0: Increase in critical bending stress due to internal pressure τ: Shear stress τR: Failure shear stress τcr: Critical shear stress τcr,0: Critical single shear stress τct,0: Critical single torsion stress τcs,0: Critical shear stress due to a single shear load τt,cr: Critical torsion stress

Contents

page V1-2.iv

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS BIBLIOGRAPHY

1-

SCHILDCROUT & STEIN Critical combination of shear and direct axial stress for curved rectangular panels. NACA T.N 1928, August 1949.

2-

G. GERARD & H. BECKER HANDBOOK OF STRUCTURAL STABILITY PART I: Buckling of flat plates. NACA T.N 3781, July 1957 PART III: Buckling of curved plates and shells. NACA T.N 3783, August 1957.

3-

ENGINEERING SCIENCES DATA UNIT Volume 1: General Data: Plasticity correction factors for plate buckling. Volume 2: Buckling of rectangular plates and panels in compression. Buckling of flat isotropic plates under uniaxial and biaxial loading. Volume 9: Cylinders and Pipes. Volume 10: Buckling of flat rectangular plates and panels in shear.

4-

BRUHN ANALYSIS & DESIGN OF FLIGHT VEHICLE STRUCTURES

5-

R.J. ROARK & W.C. YOUNG FORMULAS FOR STRESS AND STRAIN


Contents

page V1-2.v

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS



Contents

page V1-2.vi


V1-2 •1-GENERAL V1-2 •1.1-INTRODUCTION The definition and the description of the general behaviour of plates is given in Volume 2 of this manual. Simply remember that a thin plate is a prismatic body which is thin in relation to its other two dimensions. Thin plates are items with two dimensions. The buckling stress of a thin flat plate is the stress at which a change to the stable configuration of equilibrium occurs. The plate was flat then becomes slightly buckled. An increase in the external load beyond this limit cause an accelerated increase in deflection. All external loads, no matter how distributed are contained within the central plane (x,y).

z y φy

φxy

φx x The flat plate stability theory is flanged on the "thin plates with low deflection" assumptions. All components of the stress may be expressed by the deflection which is a two dimension function in the central plane. Structural items held between sheets (stiffened panel bays, spar webs, etc.) may naturally be considered as thin plates. Most angle sections may also be considered as consisting of an assembly of plates. Thus, this Chapter 2 of Volume 1 can be used to determine load buckling stresses in stringers, frames, stiffeners and other members. A shell is always thin in relation to its other dimensions and its curvature radii. The surface bisecting the thickness of a shell is the average surface. A shell is entirely defined geometrically when the shape of its average surface and its thickness at each point is specified. To analyse it, an infinitely small plate, limited by two pairs of adjacent planes, normal to the surface and containing its principal curvatures is cut from it. Revision 1 © AEROSPATIALE - 1998

General

page V1-2•1.1

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS The theory of small deformations, which gives good results for buckling flat plates in the linear elastic domain does not provide good results for buckling stresses of curved plates and shells. For curved plates and shells it is necessary to select the large displacement assumption and introduce a perturbation. This type of approach requires the use of the finite element method. This chapter of the static stressing manual presents the results of the theoretical analysis of flat plates and a semi-empirical approach for shells and curved plates.

V1-2 •1.2-SINGLE AND COMBINED LOADING The buckling stress of a plate subjected to a single load, i.e. unidirectional, may be formulated as follows: 2 eö æ σcr (or τcr) = ηKE ç ÷ è bø K, the buckling factor (linear elasticity) depends on the supporting conditions at the edges and the dimensions of the plate. The same applies to the plastic correction factor η. The plate is stable for as long as its buckling stress is greater than the stress σ (or τ) of which it φ is the seat. This stress is then linear with the applied load: σ = e To determine the buckling stresses of a plate subjected to combined loads, the interaction principle is used: A plate subjected to a system of m single loads, whereσi is the stress and σiadm,0 the buckling stress (or allowable if buckling is not possible) corresponding to the load i alone. The loading rate for load i alone is as follows: σi Ri = σ iadm,0 Interdependence of m loads is represented by an interaction equation, the general formulation of which is: R = R 1α1 + R α2 2 + ... + R αi i + ... + R αmm The exponents αi are determined by the theory or experimentally. Buckling occurs when the interaction equation result reaches 1. R is the loading rate under combined loads. R is such that: α1 αi αm æ R1 ö æ Ri ö æ Rm ö + ... + + ... + ç ÷ =1 ç ÷ ç ÷ è Rø è Rø è R ø This equation is the one of the required border. If this plate is subjected to two loads, it can be graphically illustrated in the form of an interaction curve. Revision 1 © AEROSPATIALE - 1998

General

page V1-2•1.2

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS The border is called the interaction curve and the margin with regards to buckling is: σ σ2 1 MS = - 1 R= 1 = σ 1adm σ 2 adm R

σ1adm.0

interaction curve

σ1adm σ1 Allowable loads

σ2

σ2adm

σ2adm,0

0

V1-2 •1.3-PLASTICITY CORRECTIONS In addition to the type of plate and the boundary conditions, the plastic correction factor η depends on E, Et, Es and ν. These latter values depend on the stress value to be calculated. The equation of the buckling stress can be formulated as follows:

σcr = ησcre Where σcre is the linear elastic buckling stress. Therefore, this calculation is iterative. In practice, for as long as σcre is less than the proportionality limit σe of the material, the plastic correction factor may be considered as being 1. σ For standard aluminium alloys: σe = 0.2 2 Beyond that, the Ramberg and Osgood model is used: æ σ ö σ ε = + 0.002 ç ÷ E è σ 0 .2 ø σ Es = ε 1 n 1−n = + E t Es E Revision 1 © AEROSPATIALE - 1998

n

General

page V1-2•1.3

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS The elastic-plastic Poisson ratio ν is obtained by interpolation between the elastic Poisson ratio νe and the plastic Poisson ratio νp:

ν=

Es E ö æ νe + ç1 − s ÷ νp ;νp = 0.5 (conservation of volume) è Eø E

Apply the interaction formula at single load rates corrected for plasticity, for combined loading.

V1-2 •1.4-FIELD OF APPLICATION The formulas and charts given in this chapter apply to thin plates and shells. They may be used to calculate most shell and stiffened panel items encountered in flight vehicle structures.


General

page V1-2•1.4


The 4 sides are fixed


Flat plates

page V1-2•2.37

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •2.5-EXAMPLES A - Use the example in paragraph V1-1, paragraph 2.3

y pocket x

BAY b = 170 mm e = 2 mm

a = 530 mm land

members

- Plate geometry: a = 530 mm ; b = 170 mm ; e = 2 mm - Plate loading (at L.L.): σx = 18 MPa σy = 9 MPa τxy = 27 MPa - Plate material: 2024 PL T3 E = 70300 MPa σ0.2 = 270 MPa σR = 440 MPa n = 7.05 νe = 0.33 - Plate boundary conditions: The 4 sides are considered as being hinged.


Flat plates

page V1-2•2.38

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS Critical stress in longitudinal compression (paragraph 2.4.2, page 33) - The reference stress is: 2 π2 E æ eö σ0 = ç ÷ = 36 MPa 3 (1 − ν 2e ) è b ø - Then the curves on page 34 are used: a / b = 530 / 170 = 3.1 æ σ x σ cxe ö σ cye ö æσ çç = = 2÷÷ Þ ç cxe = 2 ÷ σ0 ø è σ0 è σ y σ cye ø - It is then possible to plot curve

σ cxe a as a function of : σ0 b

The 4 sides are hinged

σ cye æ σ cxe ö = 0.5 ; = 0 ,25 ;cye = 9 MPa ç ÷ Þ σ cxe = 18 MPa σ σ0 è σ0 ø

(

)

It is not necessary to calculate the plasticity correction. Rc =

σx 18 =1 = σ cx.0 18


Flat plates

page V1-2•2.39


Critical stress in shear (paragraph 2.3.1.2, page 9)

- Rectangular plate simply supported on four edges. b - = 0.32 a 2

Giving:

b ks = 3.8 æç ö÷ + 5.35 = 5.74 è aø

π2 ks Ks = = 5.3 12 (1 − ν 2e ) 2

e τcr.0 = 5.3 E æç ö÷ ≈ 52 MPa è bø

It is not necessary to calculate the plasticity correction. Rs =

τ τ cr.0

=

27 = 0.52 52

Condition of critical stresses (paragraph 2.4.1, page 27)

According to the compression - shear interaction formula æ ö R + R 2c + 4 R s2 çR = σ = τ = c = 1.22÷ ç ÷ 2 σ cx τ cr è ø The plate folds at 82% of L.L. The folding stresses are: σx = 14.8 MPa σy = 7.4 MPa τ = 22.1 MPa


Flat plates

page V1-2•2.40

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS B - It is necessary to determine the buckling stress in longitudinal compression of the same plate, which as a constant thickness of 5 mm.

- This is the same as case 5 of paragraph 2.3.1.1, page 8. æ a = 31 . > Limö÷ Þ (kc = kc∞ = 4) ç èb ø 4 π2 = 3.69 12 (1 − 0.33 2 ) 2 σ eö æ σcce = 3.69 E ç ÷ = 224.5 MPa æç > 0.2 ö÷ è bø è 2 ø

Kc =

The table of plasticity correction factors on page 6 gives: æ 1 − ν 2e E s E ö η3 = η1 ç 0.5 + 0.25 1 + 3 t ÷ ;η1 = Es ø 1 − ν2 E è

With: [Es, Et, ν] calculated using the Ramberg and Osgood model as indicated in paragraph 1.3. This calculation is iterative: With one stress value σ1, associate η3. Then σ2 = η3 σcce Convergence is obtained when: σ1 ≈ σ2 In our case: (σ1 = 224.5 MPa)Þ (ε = 0.003738 ; sE= 60060 MPa ; tE= 31925 MPa ;ν = 0.355 ;η3 = 0.79) σ2 = 0.79 * 224.5 = 176.5 MPa σ1 + σ 2 2 σ1 = 200.5 MPa Þ η3 = 0.78 ;σ2 = 195.3 MPa σ1 = 197.9 MPa Þ η3 = 0.878 ;σ2 = 197 MPa

To ensure convergence, repeat by putting down: σ1 =

Finally: η3 = 0.879 ;σcc = 197.4 MPa It is possible to see the need to calculate the plasticity correction well before σcce = σ0.2


Flat plates

page V1-2•2.41




Flat plates

page V1-2•2.42

CASE 3

FLANGES (hinged loaded edges)

CASE 1



Flat plates

page V1-2•2.11

CASE 4

FLANGES (fixed loaded edges)

CASE 2



Flat plates

page V1-2•2.12

CASE 9

CASE 7

PLATES (hinged loaded sides)

CASE 5



Flat plates

page V1-2•2.13

CASE 10

CASE 8

PLATES (fixed loaded sides)

CASE 6



Flat plates

page V1-2•2.14

Rectangular plates - single shear

4 hinged sides 4 fixed sides



Flat plates

page V1-2•2.15

Rectangular plates - single bending

4 hinged sides 4 fixed sides



Flat plates

page V1-2•2.16

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •2.3.2-Flat parallelogram plates V1-2 •2.3.2.1-Single loads Processed cases

a b

α

α 4 fixed edges case 1

α

4 hinged edges case 2

4 fixed edges case 3

Smoothed curves, cases 1 to 3: 2 2 æç 0 ≤ a ≤ Limö÷ → æ k = C æç a ö÷ + C æç b ö÷ + C ö ç c ,s 1 2 3÷ è ø è bø è aø b è ø æ a ≥ Limö → ( k = k ) ç ÷ c ,s c ,s∞ èb ø ∏ 2 k c ,s K c ,s = 12 (1 − ν e2 )

Case 1 α=0 α=30 α=40 α=45

C1 0 0 0 0

C2 2.9 4.15 5.53 7.5

C3 6.98 6.98 6.98 6.98

Validity (a/b) 0.8 - ∞ 0.75 - ∞ 0.9 - ∞ 1.0 - ∞

Lim ∞ ∞ ∞ ∞

Kc (infinite) 6.98 6.98 6.98 6.98

Case 2 α=0 α=15 α=30 α=45 α=60

C1 1 0.5 - 0.24 - 0.45 0

C2 1 1.11 1.12 4 15.38

C3 2 2.8 5.9 7.3 4.87

Validity (a/b) 0.3 - 1.0 0.3 - 1.2 0.4 - 2.6 0.5 - 2.8 1.0 - ∞

Lim 1 1 2.6 2.8 ∞

Kc (infinite) 4 4.26 4.35 4.7 4.87

Case 3 α=0 α=45

C1 0 0

C2 5.53 4.65

C3 8.98 8.98

Validity (a/b) 0.8 - ∞ 0.8 - ∞

Lim ∞ ∞

Kc (infinite) 8.98 8.98

Note:

k (α=30) = 1.43 * k α ( =0) - 3.00 k (α=40= = 1.91 * kα=0) ( - 6.33 k (α=45) = 2.59 * l α ( =0) - 11.07

Þ Equations valid only for case 1

* All angles are in degrees ** For angles < 15°, use the curves corresponding to flat rectangular plates Revision 1 © AEROSPATIALE - 1998

Flat plates

page V1-2•2.17

Parallelogram plates with 4 fixed sides - single compression



Flat plates

page V1-2•2.18

Parallelogram plates with 4 hinged sides - single compression



Flat plates

page V1-2•2.19

Parallelogram plates with 4 fixed sides - single shear



Flat plates

page V1-2•2.20

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •2.3.3-Flat triangular plates V1-2 •2.3.3.1-Compression (with or without shear) Processed cases:

a b fixed edges case 1

hinged edges case 2

fixed edges case 3

Compression, no shear

hinged edges case 4

Compression, with shear

Smoothed curves, cases 1 to 4:

æç 0 ≤ a ≤ Limö÷ → æ k = C ç c ,s 1 è ø b è æ a ≥ Limö → ( k = k ) ç ÷ c ,s c ,s∞ èb ø ∏ 2 k c ,s K c ,s = 12 (1 − ν e2 )

CASE 1 CASE 2 CASE 3 CASE 4

C1 0 0 0 0


C2 2.8 2.09 7.73 5.11

2

æç a ö÷ + C 2 è bø

C3 9.2 2.7 28.2 8.2

2 æç b ö÷ + C ö 3÷ è aø ø

Validity (a/b) 0.5 - 2.0 0.4 - 2.0 0.5 - 1.03 0.5 - 1.3

Flat plates

Lim 2 2 1.03 1.3

Kc (infinite) 9.9 3.2 35.4 11

page V1-2•2.21

Triangular plates - compression without shear

fixed edges hinged edges



Flat plates

page V1-2•2.22

Triangular plates - compression with shear

fixed edges hinged edges



Flat plates

page V1-2•2.23

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •2.3.4-Flat plates with evolutive width V1-2 •2.3.4.1-Compression σ1: compression side b1 σ2: compression side b2 σ1

σ2

b1

æ eö 0.91 σcr = η Kc E ç ÷ 2 1 − νe è b2 ø

b2

2

CASE: σ1 = 0.8 σ2

a

fixed loaded sides, all other sides hinged

all sides are hinged

Kc

Kc

hinged loaded sides, other sides fixed

Kc

all sides are fixed

Kc


Flat plates

page V1-2•2.24

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS σ1 = 1.0 σ2

fixed loaded sides, other sides hinged


Kc

Kc

all sides are fixed


Kc


Kc

Flat plates

page V1-2•2.25

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS σ1 = 1.2 σ2

fixed loaded sides, other sides hinged


Kc

Kc


all sides are fixed

Kc


Kc

Flat plates

page V1-2•2.26

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •2.4-COMBINED LOADING Applicable to the flat rectangular plate case.

V1-2 •2.4.1-Combined loading without transverse loading The interaction curve of a plate subjected to compression combined with shear and bending is given in the following form (refer to paragraph 1.2): R αc + Rβs + Rfγ = 1

where Ri =

σi σ iadm,0

Exponents α, β, γ are determined by theory or by testing: compression + shear:

Rf = 0 ;α = 1 ;β = 2 Þ

Interaction equation: RC + RS2 = 1 2 MS = -1 R C + R 2C + 4 R S2

(Rc > 0)

Tension + shear: (tension is considered as a negative compression) Interaction equation: Rs + 0.5 Rc = 1 1 MS = -1 R s + 0.5 R c

(Rc < 0)

Bending + shear:

Rc = 0 ;β = 2 ;γ = 2 Þ

Interaction equation: Rs2 + R2f = 1 1 -1 MS = R s2 + R f2

Bending + compression:

Rs = 0 ;α = 1 ;γ = 1.75 Þ

Interaction equation: R1f.75 + Rc = 1 MS may be determined graphically using the interaction curve, on page 30 or numerically (refer to paragraph 1.2)

Bending + compression + shear: 1.75 R c + 2 R s =1 Rc + Rs MS may be determined graphically using the 1.75 R c + 2 R s α=1; β=2; γ= Þ interaction curves, on pages 31 and 32 or numerically Rc + Rs (refer to paragraph 1.2)

Interaction equation: Rc + Rs2 + Rf


Flat plates

page V1-2•2.27

Compression (tension) - shear interaction curve Rc+Rs^2=1 (if Rc>0 compression) Rs-0.5Rc=1 (if Rc<0 tension)



Flat plates

page V1-2•2.28

Shear - bending interaction curve Rs^2 + Rf^2 = 1



Flat plates

page V1-2•2.29

Compression - bending interaction curve Rc + Rf^1.75 = 1



Flat plates

page V1-2•2.30

Flat plates page V1-2•2.31



Compression - shear - bending interaction curves as a function of the different values of Rc/Rs indicated directly on the curves

Flat plates page V1-2•2.32



Compression - shear - bending interaction curves as a function of the different values of Rs/Rc indicated directly on the curves

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •2.4.2-Combined loading with transverse loading The critical stress in a flat rectangular plate is subjected to a combined loading with transverse stress (tension or compression) is calculated as follows: • Without longitudinal compression: The transverse load is processed as a longitudinal load by replacing the in value b formulas by a value. In the presence of other loadings such as shear or bending, the interaction curves in the previous paragraph are used. • With longitudinal compression: Plate stability not only depends on the various type of loads applied but also on the in-plane dimensions. In fact, the ratio (a/b) defines the buckling mode. The transverse load, whether tension or compression, affects the longitudinal critical compression stress. The curves given on the following pages show the critical longitudinal σcxe and transverse σcye stresses as a function of the a/b (length/width) ratio and the boundary conditions at plate edges. In presence of other loads such as shear or bending, the interaction curves of the previous paragraph are used, putting down: Rc =

σx where σcx = η x σcxe σ cx

To use the curves on the following page, note that σ0 is a reference stress: σ0 =

π2 E 3 (1 − ν 2e )

without longitudinal stress


æç e ö÷ è bø

2

with longitudinal stress

Flat plates

page V1-2•2.33


The 4 sides are hinged


Flat plates

page V1-2•2.34


hinged fixed

hinged


Flat plates

page V1-2•2.35


fixed hinged

fixed


Flat plates

page V1-2•2.36


V1-2 •2-FLAT PLATES z y

e

b

a

Strictly speaking, a plate is entirely bordered by tracks (for single hinging or fixed ends). The following figures show the column to plate transition when single supports (hinges) are added at unloaded edges (the loaded edges are hinged).

COLUMN

FLANGE

PLATE Revision 1 © AEROSPATIALE - 1998

Flat plates

page V1-2•2.1

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS A - SUMMARY OF CASES DEALT WITH RECTANGULAR FLAT PLATES a - Single compression (page 8, curves pages 11 to 14) FLANGES

PLATES

COLUMNS

CASE 1

CASE 5

CASE 1

CASE 2

CASE 6

CASE 2

CASE 3

CASE 7

CASE 4

CASE 8 free edge hinged edge

CASE 9

fixed edge CASE 10 b - Single shear (page 9, curves page 15)

CASE 1

CASE 2

c - Single bending (page 10, curves page 16)

CASE 1

CASE 2

CASE 3

CASE 4

Combined loading of rectangular flat planes studied on page 27, curves on pages 28 to 32. FLAT PARALLELOGRAM PLATES Single loads (page 17, curves 18 to 20)

CASE 1


CASE 2

CASE 3

Flat plates

page V1-2•2.2

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS FLAT TRIANGULAR PLATES Compression (with or without shear) (page 21, curves pages 22, 23) a b fixed edges case 1

hinged edges case 2

fixed edges case 3

Compression, no shear

hinged edges case 4

Compression, with shear

VARIABLE WIDTH FLAT PLATES Compression: (page 24, curves pages 24 to 226)

σ1

σ2

b1

b2

a

3 cases are studied:

σ1 σ σ = 0.8 ; 1 = 1.0 ; 1 = 1.2 σ2 σ2 σ2


Flat plates

page V1-2•2.3

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS B - FLAT PLATE ANALYSIS PROCEDURE Obtain the following data: - plate geometry: a, b, e - plate loading - plate material properties

Identify plate boundary conditions: free, hinged, fixed edges, etc.

Calculate the buckling factors (paragraph 2.3) Kc (as per x) if loading direction x Kc (as per y) if loading direction y Ks if shear Kf if bending

Select the plastic correction factor

Calculate the critical buckling stresses (paragraph 2.2 and 2.3) in: compression as per x compression as per y shear bending No

Is the plate subjected to combined loading?

Calculate Ri (paragraph 1.2) MS = (1/Ri) - 1

No

Yes

with transverse compression? Yes Calculate σcx (paragraph 2.4.2)

Calculate the loading rates (paragraph 2.2 and paragraph 2.4) Rc if compression Rs if shear Rf if bending

use the interaction curves (paragraph 2.4) Revision 1 © AEROSPATIALE - 1998

Flat plates

page V1-2•2.4

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •2.1-SINGLE LOADING Three single in-plane loads are possible:

a

τxy

σx

Shear

Bending

b

Compression

The corresponding buckling stresses are formulated as follows: 2

2 æ ö eö e æ σcc,0 = ηKcE ç ÷ ;τcr,0 = ηKsE ç ÷ ;σcf,0 = ηKfE è bø è min {a, b} ø

æ eö ç ÷ è bø

2

REMARK: If a plate is loaded with a transverse compression stress (as per y), replace b by a in the formula. The buckling factors of rectangular, parallelogram, triangular plates and plates with an evolutive width under single loads are given in paragraph 2.3.


Flat plates

page V1-2•2.5

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •2.2-PLASTICITY CORRECTION FACTOR The plasticity correction is applied to the buckling stresses under single loads, even if the studied plate is subjected to combined loads. Loading

Compression and bending

Boundary conditions Flange with one unloaded hinged edge

Equation æ1 − ν ö Es η1 = ç ÷ è1 − ν ø E

Flange with one unloaded fixed edge

æ E ö η2 = η1 ç 0.33 + 0.335 1 + 3 t ÷ Es ø è

Plate with unloaded hinged edges

æ E ö η3 = η1 ç 0.5 + 0.25 1 + 3 t ÷ Es ø è

Plate with unloaded fixed edges Compression

Column

Shear

All conditions

2 e 2

æ E ö η4 = η1 ç 0.352 + 0.324 1 + 3 t ÷ Es ø è Et E G η6 = s G

η5 =

In the particular case of shear, the compression stress-strain curve of the material will also be used. Proceed as follows: • Calculate the equivalent normal stress:σeq = τ 3

• Calculate the corresponding Es and ν σeq •

Gs æ 1 + νe ö Es 1− ν =ç = η1 ÷ G è 1 + ν ø E 1 − νe

Remark: η1 > η3 > η4 > η2 > η5 Choose a conservative solution if doubt subsists.


Flat plates

page V1-2•2.6

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •2.3-BUCKLING FACTORS Except in the case of columns, the plate buckling factors in the linear elastic domain K are formulated as follows: ∏2 K= k 12 (1 − ν e2 ) Classically, the curves giving k as a function of a/b are wavy. Each wave corresponds to a buckling mode, the number of half waves of the buckle pattern m increases with a/b. Owing to this, it is possible to observe relatively high variations of k for low variations of a/b. k

m=1

m=2 m=3 smoothed curve a/b

As one is never absolutely sure to choose the best boundary conditions, it is mandatory to use smoothed curves to determine k and to select conservative boundary conditions. The curves show a horizontal asymptote for a/b tending toward the infinite.


Flat plates

page V1-2•2.7

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •2.3.1-Rectangular flat plates V1-2 •2.3.1.1-Single compression Processed cases:

FLANGES

PLATES

COLUMNS

CASE 1

CASE 5

CASE 11

CASE 2

CASE 6

CASE 12

CASE 3

CASE 7

CASE 4

CASE 8

free edge hinged edge

CASE 9

fixed edge CASE 10 Smoothed curves, cases 1 to 10, except 7 and 8:

æç 0 ≤ a ≤ Limö÷ → æ k = C ç c 1 è ø b è æ a ≥ Limö → ( k = k ) ç ÷ c c∞ èb ø

2

æç a ö÷ + C 2 è bø


π2 k c 12 (1 − ν 2e ) a tends towards 0, this tends toward the column plate case. When b a case 6: if < 0.4 b Þ Use the formula for case 12 a case 10: if < 0.57 b Smoothed curves, cases 7 and 8: kc = k c (unloaded hinged edges) . k c (unloaded fixed edges) Kc =

π2 kc Kc = 12(1 − ν 2e ) Revision 1 © AEROSPATIALE - 1998

Flat plates

page V1-2•2.8

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS Smoothed curves, cases 11 and 12: 2

b kc = C2 æç ö÷ è aø 2 π2 kc π2 C2 e Kc = Þ σcc.0 = η E æç ö÷ è aø 12 12

C1 0 0 0.136 0.007 1 0 4.885 0 0 0

CASE 1 CASE 2 CASE 3 CASE 4 CASE 5 CASE 6 CASE 9 CASE 10 CASE 11 CASE 12

C2 1 4.143 0.987 3.923 1 2.7 0.965 2.9 1 4

C3 0.43 0.384 0.551 0.94 2 4 2.638 6.98 0 0

Lim ∞ 9.49 1.64 4.34 1 ∞ 0.66 ∞ ∞ ∞

kc∞ 0.43 0.43 1.28 1.28 4 4 6.98 6.98 0 0

V1-2 •2.3.1.2-Single shear

CASE 1

CASE 2

Smoothed curves: 2 æ a b ö ks = C1 ç min ìí ; üý ÷ + C 2 îb a þ ø è

ks =

Π2 ks 12 (1 − ν 2e )

4 hinged edges (case 1) 4 fixed edges (case 2)


C1 3.8 5.53

C2 = ks∞ 5.35 8.98

Flat plates

page V1-2•2.9


V1-2 •2.3.1.3-Single bending

CASE 1

CASE 2

free

CASE 3 (*) CASE 4 (*)

hinged

fixed

Smoothed curves: 2 æç 0 ≤ a ≤ Limö÷ → æ k = C æç a ö÷ + C 1 2 è ø çè f è bø b æ a ≥ Limö → k = k ç ÷ ( f f∞ ) èb ø π2 k f Kf = 12 (1 − ν 2e )

C1 CASE1 3.33 CASE2 52.3 CASE3 CASE4

C2 1.66 2.2

C3 18.1 20.4


Validity (a/b) 0.4 - 0.6 0.3 - 0.45

Lim 0.6 0.45 1.3 1.3

kf∞ 23.9 41.8 1.96 0.85

(*) the free edges are compressed no reliable curve in cases 3 and 4 for a/b < 1.3


Flat plates

page V1-2•2.10


Buckling factor in shear for long curved panels, simply supported


Curved plates

page V1-2•3.11


Buckling factor in shear for wide curved panels, simply supported


Curved plates

page V1-2•3.12

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •3.4-COMBINED LOADING V1-2 •3.4.1-Shear + compression (or tension) The interaction curve of a curved plate subjected to axial compression combined with shear is given in the following form: σ Interaction equation: Rc + Rs2 = 1 with Rc = σ cc,0 2 τ Þ MS = -1 Rs = 2 2 τ cr,0 Rc + Rc + 4 Rs where: Rc > 0 Þ compression

and

Rc < 0 Þ tension

V1-2 •3.4.2-Compression - side pressure The existence of internal pressure p increases the axial buckling stress in compression of the curved sheet panel σcc. The presence of an external pressure reduces it. Here, side pressure is present and not hydrostatic pressure: pr • Side pressure: σx = 0 ;σy = e pr pr • Hydrostatic pressure: σx = ;σy = 2e e 2 The interaction equation is: Rc + Rp = 1 σ in which Rc = c σ cc Rp designating the ratio between the pressure effectively applied and the external pressure that would have to be applied to buckle a cylinder of the same radius as the panel (refer to paragraph 4.3.1.5, on page 4/15). This external pressure is calculated using the curve on page 14. In the above interaction equation, the value Rp is taken as negative in the internal pressure case.

V1-2 •3.4.3-Shear + side pressure The existence of an internal pressure p increases the buckling stress in shear of a curved sheet panel τcr. The presence of an external pressure reduces it. Here, side pressure is present and not hydrostatic pressure: pr • Side pressure: σx = 0 ;σy = e pr pr • Hydrostatic pressure: σx = ;σy = 2e e 2 The interaction equation is: Rs + Rp = 1 (Rp < 0 in the case of internal pressure) τ in which Rs = τ cr Rp designating the ratio between the pressure effectively applied and the external pressure which would have to be applied to buckle a cylinder of the same curvature radius as the panel (refer to paragraph 4.3.1.5, page 4/15). This external pressure will be calculated using the curve on page 14. Revision 1 © AEROSPATIALE - 1998

Curved plates

page V1-2•3.13

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS Buckling factor under external radial pressure ky as a function of Z

radial pressure

with: r 1 − ν 2e e 1 a2 2 2 Z = (1 − ν e ) re k y π2 E σcy = 12 (1 − ν 2e ) t=


æç e ö÷ è aø

2

Curved plates

page V1-2•3.14

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •3.5-EXAMPLES A - Use the example in V1-1, paragraph 2.3.

Assume that the plate is long and curved with a curvature radius of r.

y pocket x

BAY

b = 170 mm e = 2 mm

a=530 mm land

- Curved plate geometry:

- Plate loading (at L.L.):

members

a = 530 mm b = 170 mm e = 2 mm r = 2800 mm σx = 18 MPa

- Plate material: E = 70300 MPa σ0.2 = 270 MPa σR = 440 MPa n = 7.05 νe = 0.33 - Plate boundary conditions: The 4 sides are considered as being hinged.


Curved plates

page V1-2•3.15

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS Longitudinal compression:

- Calculation of the buckling factor (using figure page 8): Z=

1 b2 1 − ν 2e ) 2 = 4,87 ( re

Kc = 4.95 r/e = 2800 / 2 = 1400 Thus, the critical longitudinal stress is (refer to paragraph 3.3). σcc,0 = η

Kc π 2 E 12 (1 − ν 2e )

2

æç e ö÷ = η x 44.5 MPa è bø

Obviously it is not necessary to calculate the plasticity correction (η = 1). σcc,0 = 44.5 MPa

Rc =

σ σ cc, 0

=

18 = 0.4 44.5

æ 1 ö The margin is: MS = ç ÷ - 1 = 1.5 Þ folding at 250% of L.L. è Rc ø B - Use the example above assuming that in addition to longitudinal compression, the curved plate is subjected to a 0.01MPa (refer to paragraph 3.4.2) internal pressure.

The buckling stress in compression was calculated above: σcc,0 = 44.5 MPa To assess Rp, it is necessary to determine the external radial pressure which would cause buckling of a circular cylinder with the same curvature as the curved sheet panel: (curves, page 14). 530 2 a2 2 Z= 1 − νe = 1 −0.33 2 = 47.4 2800 x 2 re Ky π 2 E

2 æç e ö÷ = 8.5 x π x 70300 x æç 2 ö÷ = 7.85 MPa ky = 8.5 Þ σcr = 12 (1 − ν 2e ) è a ø 12 x (1 − 0.33 2 ) è 530 ø 2

2

The external radial pressure capable of producing this buckling is: Pcr = σcr

e 2 = 7.85 x = 0.0056 MPa r 2800

Þ

Rp = -

0.01 = - 1.78 0.0056

The interaction equation gives (refer to paragraph 1.2) σ 2cc + Rp = 1 σ 2cc.0

being

σ 2cc - 1.78 = 1 44.5 2

Givingσcc = 74.2 MPa

Thus, the effect of a 0.01 MPa internal pressure is to increase the axial buckling stress from 44.5 MPa to 74.2 MPa. Revision 1 © AEROSPATIALE - 1998

Curved plates

page V1-2•3.16

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS C - Use the example in A above assuming that in addition to longitudinal compression, the curved plate is subjected to a 27 Mpa shear. Longitudinal compression: (already calculated above)

σcc,0 = 44.5 MPa

Rc =

σ σ cc, 0

=

18 = 0.4 44.5

Shear: (as per paragraphs 3.1 and 3.3.1.2)

Ks π 2 E Linear elastic calculation: τcr,0 = η 12 (1 − ν 2 )

æç e ö÷ è bø

2

According to the curves on page 11: 1 a 530 b2 Z = (1 − ν 2e ) 2 = 4.87 and æç = = 31 . ö÷ Þ (Ks = 6.2) è b 170 ø re Thus: Ks π 2 E æ e ö 2 τcr,0 = η x ç ÷ = η x 55.7 MPa 12 (1 − ν 2e ) è b ø It is obviously not necessary to calculate the plasticity correction. τcr,0 = 55.7 MPa

Rs =

τ τ cr, 0

=

27 = 0.48 55.7

Compression - shear interaction

æ ö R + R 2c + 4 R s2 çR = σ = τ = c = 0.72÷ Þ Folding 139% of L.L. ç ÷ 2 σ cx τ cr è ø


Curved plates

page V1-2•3.17

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS D - Use the same example A above but with a 5 mm pocket thickness.

Critical stress in compression - Calculation of the buckling factor (using figure page 8): 1 b2 Z = (1 − ν 2e ) 2 = 1.95 re Kc = 4.19 r 2800 = = 560 e 5 Thus, the critical longitudinal stress is (refer to paragraph 3.3). Kc π 2 E æ e ö 2 σcc,0 = η ç ÷ = η x 235.2 MPa 12 (1 − ν 2e ) è b ø - Calculation of the plasticity correction factor, iterative method: Use the Ramberg and Osgood formulas (refer to paragraph 1.3) at the first step of the calculation: n 7. 05 æ σ ö 235.2 ö σ 235.2 æ + 0.002 ç ε = + 0.002 ç = 0.0041 ÷ ÷ = è 270 ø 70300 E è σ 0 .2 ø E E ö σ 235.2 æ = = 57366 MPa and ν = s x 0.33 + ç1 − s ÷ x 0.5 = 0.361 è ε 0.0041 E Eø 1 n 1− n 7.05 1 − 7.05 = + = + = 0.00003684 Þ Et = 27148 MPa E t Es E 57366 70300

Es =

According to paragraph 3.2, page 3: ìïæ 1 − ν 2e ö E s æ E t ö üï Es ç ÷ x x . + . + + c x x 0 5 0 2 1 3 η3 = (1 - c) x íç ÷ ý E s ø ïþ E ïîè 1 − ν 2 ø E è 1 æ b2 ö − 1÷ = 0.01075 with c = ç ø 99 è re 0.5 + 0.25 1 + 3

E Et = 0.889 and s x E Es

Et 57366 = x E s 70300

Et Es

27148 = 0.561 57366

ìæ 1 − 0.33 2 ö 57366 ü x x (0.889)ý + 0.01075 x 0.561 = 0.74 η = (1 - 0.01075) x íç 2÷ îè 1 − 0.361 ø 70300 þ

The critical stress σcc,0 is then: 0.74 * 235.2 = 174 MPa To ensure convergence, it is necessary to repeat the operation using as an initial value for each iteration, the average between the value found and the initial value of the previous step σ = (235.2 + 174) / 2 = 204.6 MPa σ = 204.6 MPaÞ ε = 0.00319 Þ Es = 64070 MPa Þ Et = 41707 MPa Þ η = 0.856 Þ σcc.0 = 201.3 MPa σ = 203 MPa Þ ε = 0.00316 Þ Es = 64334 MPa Þ Et = 42509 MPa Þ η = 0.861 Þ σcc.0 = 202.5 MPa

Thus, after several iterations, convergence is towards σ 18 η = 0.862 and σcc,0 = 202.7 MPa Þ Rc = = 0.089 = σ cc, 0 202.7 Revision 1 © AEROSPATIALE - 1998

Curved plates

page V1-2•3.18

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS - Calculation of the plasticity correction factor, schematic method: Plot the curve giving σcr corrected according to elastic σcr: Assume as an initial value: σcr corrected =

σ 0.2 270 = = 135 MPa 2 2

n

7. 05 æ σ ö σ 135 135 ö + 0.002 æç = 0.00194 ε = + 0.002 ç ÷ ÷ = è 270 ø 70300 E è σ 0 .2 ø σ 135 = 69752 MPa Es = = ε 0.00194

æ n 1 − nö Et = ç + ÷ E ø è Es

−1

7.05 1 − 7.05 ö = æç + ÷ è 69752 70300 ø

−1

= 66609 MPa

Es E ö 69752 69752 ö æ νe + ç1 − s ÷ νp = x 0.33 + æç1 − ÷ x 0.5 = 0.331 è è ø 70300 70300 ø E E æ 1 − ν 2e ö E s æ E ö E Et = 0.985 η = (1 - c) ç ç 0.5 + 025 1 + 3 s ÷ + c t 2÷ Eø Es Es è1 − ν ø E è

ν=

σ elastic =

σ 135 = 137.1 MPa = η 0.985

Thus, for σ corrected = 135 Mpa Þ σ elastic = 137.1 MPa By varying the corrected σcr, it is possible to calculate the corresponding elastic stress making it possible to plot the curve σcr corrected according to elastic σcr (refer to the next page). Then the required result is obtained directly from this curve:

σ elastic = 235.2 MPa Þ σ corrected = 202.5 MPa Þ Rc =

σ σ cc, 0

=


18 = 0.089 202.5

Curved plates

page V1-2•3.19


Elastic stress

Plasticity correction

Corrected stress


Curved plates

page V1-2•3.20


V1-2 •3-CURVED PLATES z y

e b

a r

The curved plates discussed in this chapter are rectangular plates with a single curvature in the loaded edge plane.


Curved plates

page V1-2•3.1

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS A - SUMMARY OF CASES DEALT WITH RECTANGULAR CURVED PLATES a - Single compression, single shear (page 7, curves pages 8 to 12)

hinged edge Case 1

Case 2 fixed edge

Case 3


Case 4

Curved plates

page V1-2•3.2

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS B - CURVED PLATE ANALYSIS PROCEDURE Obtain the following data: - plate geometry: - plate loading - plate material properties Identify plate boundary conditions: free, hinged, fixed edges, etc.

Determine the buckling factors K (paragraph 3.3) Kc (in x direction) if stress in x direction Ks if shear Select a plastic correction factor (paragraph 3.2)

Calculate the elastic-plastic buckling critical stresses (paragraphs 3.2 and 3.3) in: compression in x direction shear No

Is the plate subjected to combined loading?

Calculate Ri (paragraph 1.2)

Yes

Calculate the loading rates (paragraphs 1.2 and 3.4) Rc if compression Rs if shear Use the interaction equation to calculate R and MS (paragraph 3.4)


Curved plates

page V1-2•3.3

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •3.1-SINGLE LOADING Two single stresses are possible: τxy x b

a

y

Shear

Compression

The corresponding buckling stresses are formulated as below: 2 æ ö e e σcc,0 = ηKcE æç ö÷ ;τcr,0 = ηKsE ç ÷ è bø è min{a, b} ø

2

REMARKS: The buckling stress equation is the same as that for flat plates with higher buckling factors Kc and Ks. The buckling factors of rectangular curved plates are given in paragraph 3.3.


Curved plates

page V1-2•3.4

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •3.2-PLASTICITY CORRECTION FACTOR The plasticity correction is applied to buckling stresses under single load even if the studied plated is subjected to combined loads. Loading

Boundary Flat plate conditions Flange, one æ 1 − ν 2e ö E s ÷ unloaded edge η1p = ç è 1 − ν2 ø E hinged Plate with E ö æ hinged η3p = η1p ç 0.5 + 0.25 1 + 3 t ÷ Compression unloaded Es ø è edges Plate with E ö æ fixed unloaded η4p = η1p ç 0.352 + 0.324 1 + 3 t ÷ Es ø è edges Shear All conditions G η= s G

Cylinder ηc =

Es

Et

E

Es

ηc =

Es

Et

E

Es

ηc =

Es

Et

E

Es

η=

Gs G

Curved plate η1 = (1 - c)η1p + cηc

η3 = (1 - c)η3p + cηc

η4 = (1 - c)η4p + cηc η=

Gs G

1 æ b2 ö Factor c depends on the geometry of the curved plate: c = − 1÷ ç ø 99 è re The plasticity correction coefficients for a curved plate are calculated by linear interpolation b2 between the coefficients of the flat plate given with a geometric limitation 1 > and those re b2 > 100. of the cylinder given for re Thus, the geometric boundarys of the plasticity correction coefficients given in the table above are: b2 ≤ 1 : η = ηp (flat plate) re b2 ≥ 100 : η = ηc (cylinder) re b2 1< < 100 : η = ηi (curved plate) re The stress-strain compression curve is also used in the particular case of shear loading. Proceed as follows: • Calculate the equivalent normal stress:σeq = τ 3 • Calculate Es and ν corresponding to σeq. G æ 1 + νe ö Es 1− ν • s =ç η1 = ÷ G è 1 + ν ø E 1 − νe

Remark: η1 > η3 > η4 Select a conservative solution if any doubt subsists. Revision 1 © AEROSPATIALE - 1998

Curved plates

page V1-2•3.5

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •3.3-BUCKLING FACTORS The plate buckling factors K in the linear elastic domain are formulated as follows: K=

π2 k 12 (1 − ν 2e )

1 b2 2 2 1 − ν ( e ) (factor depending on the re geometry and on the material of the plate) are given in a logarithmic scale.

The curves representing k as a function of Z =

Note that: - The k factors corresponding to small values of Z (Z < 1), therefore high values of r must satisfy the flat plate theory. - The k factors corresponding to high values of Z, therefore small values of r, must satisfy the cylindrical shell theory.


Curved plates

page V1-2•3.6

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •3.3.1-Rectangular curved plates V1-2 •3.3.1.1-Single compression Processed cases:

hinged edge Case 3

Case 1

fixed edge

The curves for long curved panels representing the kc buckling factor as a function of Z are given on the next page. The linear theory for small deflections applicable in the case of single compression of flat plates, provides theoretical results higher than the results from tests performed on curved plates. Use the recommended curves to determine k.

V1-2 •3.3.1.2-Single shear Processed cases:

hinged edge fixed edge Case 2

Case 4

For each case, the long panels and the wide panels are processed. The curves representing the buckling factor ks in function of Z are given on pages 9 to 12.


Curved plates

page V1-2•3.7

Curved plates

4 hinged edges 4 fixed edges

Buckling factor in axial compression for long curved plates (*) *4 hinged sides:

page V1-2•3.8

4 fixed sides:



Recommended curves, Reference N.A.C.A. T.N. 1928

Theoretical curve

Buckling factor in shear for fixed long curved panels



Curved plates

page V1-2•3.9


cylinder

Buckling factor in shear for fixed wide curved panels


Curved plates

page V1-2•3.10

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •4.4.2-Conical shells The interaction equation relating the compression, the bending and the shear load and the external pressure is: (Rc + Rf + Rcs)1.2 + R1p.2 = 1 with: Rc =

compression stress critical compression stress

Rf =

bending stress critical bending stress

Rcs =

shear stress due to a shear load 1.4 * critical torsion stress

Rp =

applied pressure (internal pressure: Rp < 0) critical pressure

V1-2 •4.5-EXAMPLES A- Take a circular cylinder made of 2024 PL T3 aluminium alloy with the following geometrical and mechanical properties: - Cylinder geometry: L = 1950 mm r = 1300 mm e = 1,3 mm - Cylinder material: 2024 PL T3 E = 70300 MPa σ0.2 = 270 MPa σR = 440 MPa n = 7.05 νe = 0.33 - Cylinder boundary conditions: The ends are hinged - Cylinder boundary conditions: Compression load = 135 000 N Bending moment = 156 000 m.N Torsion moment = 265 000 m.N Internal pressure = 350 hPa Revision 1 © AEROSPATIALE - 1998

Cylinders and shells

page V1-2•4.33

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS Longitudinal compression with internal pressure (according to paragraphs 4.3.1.1 and 4.4.1.B): - Calculation of the buckling factor: r 1300 L 1950 = = 1000 ; = = 1.5 e 1.3 r 1300 1950 2 L2 Z= 1 − ν 2e = * 1 − 0.33 2 = 2124 1300 * 1.3 re Consultation of curves, page 10 (or formulas, page 9) Þ C = 0.171 and kc = 421 k c0 1 = = 10 << 2124Þ long cylinder Z* = 0.58 * C 0.58 * 0.171 e σcc,0 = η C E æç ö÷ = η * 0.171 * 70300 * è rø

æ 1.3 ö = η * 12 MPa ç ÷ è 1300 ø

Shows that it is obviously not necessary to calculate the plasticity correction (η = 1) The critical axial load is (without taking internal pressure into account): Pcr = 2 π r e σcc,0 = 127 423 N The internal pressure improves the critical compression load: according to page 30: 2 2 æ p ö æ r ö = æ 0.035 ö æ 1300 ö = 0.5 ç ÷ç ÷ ÷ ç ÷ç è E ø è eø è 70300 ø è 1.3 ø according to the figure on page 30: ∆σ cc,0 æ r ö 70300 * 1.3 = η = 12.3 MPa ç ÷ = 0.175 Þ ∆σcc,0 = η * 0.175 * η è E eø 1300 The global critical compression stress: σ σcc,0 + ∆σcc,0 = 24.3 MPa << 0.2 giving η = 1 2 On the other hand, the internal pressure induces a longitudinal tension load (page 30). σt =

pr 0.035 * 1300 = 17.5 MPa Þ Ft = 2 π r e σt = 185 825 N = 2e 2 * 1.3

An external compression load should have already cancelled out this tension load before subjecting the cylinder to a longitudinal compression. The critical compression load is then: Pc,cr = 2 π r e (σcc,0 + ∆σcc,0 + σt) = 2 π * 1300 * 1.3 * (12 + 12.3 + 17.5) = 443 857 N The margin is given by the following formula: M.S =

Pc,cr p

−1=

443 857 - 1 = 2.29 135 000



page V1-2•4.34

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS Bending moment with internal pressure: - Calculation of the buckling factor: r L = 1000 , = 1.5 , Z = 2124 e r Consultation of the curve, page 12 (or formulas, page 11) Þ Kf = 553.5 k c0 1 = = 7.8 << 2124Þ long cylinder Cf = 0.222 Þ Z* = 0.58 * C f 0.58 * 0.222 e σcf,0 = η Cf E æç ö÷ = η * 0.222 * 70300 * è rø

æ 1.3 ö = 15.6 MPa (η = 1) ç ÷ è 1300 ø

Thus, the critical bending moment in the absence of internal pressure is: I Mf,cr = σcf,0 σcf,0 * π * r² * e = 15.6 * π * 1300² * 1.3 = 107 672 m.N r The internal pressure improves the critical bending moment: according to page 31: 2 2 æ p ö æ r ö = æ 0.035 ö æ 1300 ö = 0.5 ÷ ç ÷ç ÷ ç ÷ç è E ø è eø è 70300 ø è 1.3 ø according to the figure on page 31: ∆σ cf ,0 æ r ö 70300 * 1.3 = η = 22.1 MPa ç ÷ = 0.315 Þ ∆σcf,0 = η * 0.315 * E è eø 1300 The global critical bending stress:

σ σcf,0 + ∆σcf,0 = 15.6 + 22.1 = 37.7 MPa << 0.2 giving η = 1 2

On the other hand, the internal pressure induces a longitudinal tension load (page 31). σt =

pr 0.035 * 1300 = 17.5 MPa = 2e 2 * 1.3

The critical bending moment is then: Mf,cr = π r² e (σcf,0 + ∆σcf,0 + σt) = π * 1300² * 1.3 * (15.6 + 22.1 + 17.5) = 381 000 m.N The margin is given by the following formula: M=

M f ,cr Mf

−1=

381 000 - 1 = 1.44 156 000



page V1-2•4.35

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS Torsion moment with internal pressure:

- Calculation of the buckling factor: r = 1000 , e

Z = 2124

Intermediate cylinder with hinged edges (refer to page 13). 3

kt = 0.85 * Z4 = 0.85 * (2124)0.75 = 266 (or according to the figure on page 14) Kt =

π2 kt = 245.5 12 (1 − 0.33 2 ) 2

2

e 1.3 ö τct,0 = η Kt E æç ö÷ = η * 245.5 * 73800 * æç ÷ = 8.1 MPa (η = 1) è Lø è 1950 ø

The critical torsion moment is: Mt,cr = τct,0

J = τct,0 * 2π r² e = 8.1 * 2π * 1300² * 1.3 = 111 814 m.N r

Internal pressure improves the critical torsion moment. The critical hydrostatic pressure pcr, causing the collapse of the cylinder is according to the figure on page 17 (or formulas, page 16): Z = 2124 Þ kp = 40.2 Þ Kp = kp *

π2 = 37.1 12 (1 − 0.33 2 )

2

2

e 1.3 ö σcp,0 = η * Kp * E * æç ö÷ = η * 37.1 * 73800 * æç ÷ = 1.2 MPa (η = 1) è Lø è 1950 ø

The critical pressure causing this circumferential stress in the cylinder is: e 1.3 Pcr = σcp,0 = 1.2 * = 12 hPa r 1300 The critical torsion stress with internal pressure is, according to the figure on page 32: Pint - Pcr = 350 - 12 = 338 hPa τct,0 = 38 MPa The critical torsion moment is then: Mt,cr = τct,0 * 2π r² e = 38 * 2π * 1300² * 1.3 = 524 558 m.N The margin is given by the following formula: M=

M t ,cr Mt

−1=

524 558 - 1 = 0.98 265 000



page V1-2•4.36

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS Compression - bending - torsion interaction

The applied loads are: - P = 135 * 103 N (compression load) - Mf = 156 * 103 m.N (bending moment) - Mt = 265 * 103 m.N (torsion moment) - Pint = 350 hPa (internal pressure) The critical loads calculated taking internal pressure into account: Pc,cr = 443 857 N Mf,cr = 381 000 m.N Mt,cr = 524 558 m.N Giving: Rc =

P 135 000 = = 0.3 Pc,cr 443 857

Rf =

Mf 156 000 = = 0.4 M f ,cr 381 000

Rct =

Mt 265 000 = = 0.5 M t ,cr 524 558

The interaction equation is: Rc + Rf + R2ct = 1 0.3 + 0.4 + 0.5² = 0.95 < 1 The cylinder does not collapse under this combined load.



page V1-2•4.37





page V1-2•4.38

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •4.3.1.2-Single bending The critical buckling stress under single bending is formulated as below: 2 e σcf,0 = η Kf E æç ö÷ è Lø The buckling factor Kf is formulated as below: π2 Kf = kf 12 (1 − ν 2e ) Calculation of kf The buckling factor depends on the parameter: Z =

L2 1 − ν 2e re

* Short cylinders and transition range kf = kc0 +

(0.581 x C f x Z) 2 k c0

Kc0 is the buckling factor of column plates under compression: fixed edges: kc0 = 4 hinged edges: kc0 = 1 * long cylinders kf = 1.16 x Cf x Z e The critical buckling stress becomes: σcf,0 = η Cf E æç ö÷ è rø

à Cf factor: The Cf factor above is: Cf = 1.3 * C (factor used in compression) −0 .15 æ ö r C = 0.6 x ln ç 3.748 x æç ö÷ ÷ ; C≤ 0.6 è eø è ø

valid for: 8.52 < if

r < 3000 e

r < 8.52 take C = 0.6 e

à Border between the transition range and long cylinders: k c0 Z* = and kf* = 2 kc0 0.58 C f Revision 1 © AEROSPATIALE - 1998


page V1-2•4.11

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS Buckling factor in bending, kf, as a function of Z loaded sides fixed



loaded sides simply supported

page V1-2•4.12

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •4.3.1.3-Single torsion Unlike what happens in compression, cylinders subjected to torsion do not spontaneously collapse after the occurrence of folding. If a cylinder is gradually loaded, first of all slight deflections occur increasing in practically a linear manner. The curvature gradually flattens around the occurrence of folding then major deflections occur for a slight variation in the load. The torsion moment drops when the cylinder collapses, whereas the side deflections continue to increase. Collapsing load The folding stress is approximately 84% of the collapsing stress.

The critical collapsing stress under torsion is formulated as follows: 2 eö π2 æ τct,0 = η Kt E ç ÷ with Kt = kt è Lø 12 (1 − ν 2e ) Calculation of Kt L2 The buckling factor depends on parameter: Z = 1 − ν 2e re

* Short cylinders For Z ---> 0: the cylinder behaves like an infinitely long flat plate. Short cylinders have as a Z factor: 0 < Z < 1001 − ν 2e - fixed edges: kt = 8.98 + 0.204 Z - hinged edges: kt = 5.35 + 0.216 Z * Intermediate cylinders 3

r2 - fixed edges: 100 1 − ν < Z < 63 2 1 − ν 2e and kt = 0.93Z4 e 3 r2 - hinged edges: 100 1 − ν 2e < Z < 44 2 1 − ν e2 and kt = 0.85Z4 e 2 e

* long cylinders r2 - fixed edges: Z > 63 2 e

- hinged edges: Z > 44

1 − ν e2 r2 e2

kt =

1 − ν e2

0.33 Z 1

æç r 1 − ν 2 ö÷ 2 e èe ø

τcr is then formulated in the simpler form: τcr = η x 0.272 (1 − ν

2 e

)


−3 4

3

e 2 E æç ö÷ è rø


page V1-2•4.13

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS Buckling factor under torsion, kt, as a function of Z

fixed hinged

V1-2 •4.3.1.4-Single shear Collapsing load

The critical collapsing stress under a shear load is formulated as follows: e π2 kt τcs,0 = 1.25 η Kt E æç ö÷ with Kt = è Lø 12 (1 − ν 2e ) 2

Calculation of kt

kt is the buckling factor calculated in the torsion case studied in the previous paragraph.



page V1-2•4.14

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •4.3.1.5-External radial pressure Collapsing load

The critical collapsing stress under external radial pressure is formulated as follows: 2 e π2 ky σcy,0 = η Ky E æç ö÷ with Ky = è Lø 12 (1 − ν 2e ) Remember that with a radial pressure p, the longitudinal stress is nil whereas the transverse stress is equal to pr/e. Calculation of Ky (chart, page 17) L2 The buckling factor depends on the parameter: Z = 1 − ν 2e re * Very short cylinders

The case of very short cylinders is identical to that of infinitely long flat plates in axial compression. Values for Z=0: - hinged edges: ky = 4 - fixed edges: ky = 6.98 * Short cylinders and long cylinders The difference between these cylinders is the number of waves obtained in the circumferential direction: short cylinders: the number of waves is greater than 2. long cylinders: only 2 waves. Let n0, be the number of circumferential waves obtained in the specific case r where Z = t, with: t = 1 − ν 2e ; Z = t⇔ L = r e The ky factor calculation procedure (then σcr) is as follows: (refer to the Appendix).

n 20 = 9.3 Z0.5 n2 =

t 2 n0 Z

n2 < 4 take n² = 4 4Z β2 = 2 π t

β2 =

(1 + β ) =

2 2

ky

β2

+

n 20 π2

12 Z2

π 4 β 2 (1 + β 2 )

2

π2 k y E æ e ö 2 σcr = ç ÷ 12 (1 − ν 2e ) è L ø Revision 1 © AEROSPATIALE - 1998


page V1-2•4.15

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •4.3.1.6-Hydrostatic pressure Collapsing load

The critical circumferential collapsing stress under hydrostatic pressure is formulated as follows: e π kp σcy,0 = η Kp E æç ö÷ with Kp = è Lø 12 (1 − ν 2e ) 2

2

Remember that with hydrostatic pressure p, the axial stress is equal to pr/2e, whereas the transverse stress is equal to pr/e. Calculation of kp:

The buckling factor depends on the parameter: Z =

L2 re

1 − ν 2e

* Very short cylinders The case of very short cylinders is identical to that of infinitely long flat plates in axial compression. Values for Z=0: - hinged edges: kp = 2 * Short cylinders and long cylinders The behaviour of long cylinders under hydrostatic pressure is similar to that of long cylinders under radial pressure. This gives a kp curve that overlays the one obtained under radial pressure for very high Z values which converge for small Z values towards kp=2 instead of 4. The kp curve will be conservatively plotted in the form of two straight lines (logarithmic scale): kp = 2

for 0 < Z < 10

kp = 0.5514 Z0.56

for


Z > 10


page V1-2•4.16

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS Buckling factor under external radial pressure, ky, as a function of Z

radial pressure

hydrostatic pressure



page V1-2•4.17

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •4.3.2-Elliptical cylinders V1-2 •4.3.2.1-Single compression Elliptical cylinders behave in the same manner as circular cylinders under the effect of compression. Therefore, the same formulas and the same charts as those used for the study of circular cylinders will be used but this time with a curvature radius equal to a²/b instead of r (circular cylinder radius).

V1-2 •4.3.2.2-Single bending Calculation method I 1- Calculate the strength modulus of the circumscribed circle: Sa = = π a² e ν σ a elliptical cross section e Compression

a r

Ycr

σcr Tension

b

circumscribed circle

2- Calculate the strength modulus of the elliptical cylinder cross section Se = (Se/Sa)*Sa S where e is the cross section efficiency. Sa

Equation of the curve:


2 Se b b = 0.424 + 0.218 *æç ö÷ + 0.521 * æç ö÷ - 0.163 * è aø è aø Sa


æ bö ç ÷ è aø

3

page V1-2•4.18

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS r y 3- Calculate the following ratios: æç ö÷ and ( r : critical curvature radius) è a ø cr a 2 æç y ö÷ = 0.5 * é1 − æç b ö÷ ù ê è ø ú è a ø cr a û ë

if

−0 . 5

æ rö a b and ç ÷ = 0.649 * æç ö÷ with 0 ≤ ≤ 0.866 è bø a è aø

b > 0.866 buckling occurs at the end of the long axis Þ r = a a

4- Calculate the stress applied at the critical curvature: M y σycr = f * æç ö÷ ; Mf is the bending moment. S e è a ø cr 5- Calculate the allowable stress (long elliptical cylinders): −0 .15 é ù C f * E *e æ rö σcr = η with Cf = 1.3 * 0.6 * In ê3.748 * ç ÷ ú r è eø êë úû 6- Calculate the margin between the allowable stress and the applied stress. Revision 1 © AEROSPATIALE - 1998


page V1-2•4.19

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •4.3.2.3-Single torsion Collapsing load The torsion moment applied around the elliptical section of the cylinder first of all causes folds at the small axis, moving gradually towards the long axis where collapsing takes place. As the curvature radius at the long axis is the smallest, this area stabilises the cylinder until the folds cause collapsing.

b

a

folding propagation direction The critical collapsing stress is calculated as for axi-symmetrical cylinders, putting down r = a (refer to paragraph 4.3.1.3).

V1-2 •4.3.2.4-Single shear Collapsing load The critical collapsing stress under a shear load is formulated as follows: e τcs,0 = 1.25 η Kt E æç ö÷ è Lø

2

π2 with Kt = kt 12 (1 − ν 2e )

Calculation of kt:

kt is the buckling factor calculated for the torsion case.



page V1-2•4.20

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •4.3.3-Spherical shells V1-2 •4.3.3.1-External pressure The critical buckling stress under external pressure is formulated as below: 2 eö æ σcp,0 = η Κp Ε ç ÷ è dø The buckling factors Kp are formulated as follows: ∏2 Kp = kp 12 (1 − ν 2e ) Calculation of kp

The buckling factor depends on the parameter: Z =

d2 re

1 − ν 2e

* Spherical shells with small Z (0.581 x C x Z) 2 k pl kpl is the buckling factor of flat circular plates. kpl = 3 kp = kpl +

* Spherical shells with long Z

e r d

kp = 1.16 x C x Z e The critical buckling stress becomes: σcp,0 = η C E æç ö÷ è rø

à C factor:

The C factor used in the formulas above is: C = eA with A = - 0.199 * In (r/e) - 0.235 r valid for: 10 < < 1000 e r if > 1000 take C = 0.2 e à Border between two curves

Z* =

k pl 0.58 C

and kp* = 2 kpl with kpl = 3



page V1-2•4.21


kp as a function of Z



page V1-2•4.22

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •4.3.4-Conical shells V1-2 •4.3.4.1-Single compression The critical buckling stress under compression is formulated as: Pc,cr = 2 π r e σc,cr Pc,cr: critical compression load. σc,cr: critical compression stress (calculation, page 24). ρ: minimum curvature radius. α: cone half angle. r: cone small radius. e: cone thickness.

Pc,cr r ρ α

L

Pc,cr Mf,cr r

V1-2 •4.3.4.2-Single bending The critical buckling stress under bending is formulated: Mf,cr = π r² e σf,cr Mf,cr: critical bending moment. σf,,cr: critical bending stress (calculation, page 25).

ρ α

L

Mf,cr

V1-2 •4.3.4.3-Single torsion Mt,cr r

The critical buckling stress under bending is formulated: Mt,cr = 2 π r² e τt,cr Mt,cr: critical torsion moment. τt,,cr: critical torsion stress (calculation, page 26).

ρ α

L

V1-2 •4.3.4.4-Single external pressure

Mt,cr

The critical external pressure is given by the following formula: pcr =

α

π2 e3 k y 12 (1 − ν 2e ) ρ moy L2

r

pcr: critical pressure (calculation, page 27).

Pcr

L ρmoy

ky: buckling factor under external pressure. R ρmoy: average curvature radius =


r+R 2 cos α


page V1-2•4.23

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •4.3.4.5-Single loading curves a) in compression critical compression stress

Curve equations: curves valid in the elastic domain with: 5° < α < 11° L 1000 ρ = 4 : σc,cr * ( ) = 4220 * æç ö÷ è eø ρ E

−1. 575

L 1000 ρ = 2 : σc,cr * ( ) = 5040 * æç ö÷ è eø ρ E

−1. 575

L 1000 ρ = 1 : σc,cr * ( ) = 5851 * æç ö÷ è eø ρ E

−1. 575

L 1000 ρ = 0.5 : σc,cr * ( ) = 6569 * æç ö÷ è eø ρ E Revision 1 © AEROSPATIALE - 1998

−1. 575


page V1-2•4.24

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS b) in bending critical bending stress

Curve equations: curves valid in the elastic domain with: 5° < α < 11° L 1000 ρ = 4 : σf,cr * ( ) = 5361 * æç ö÷ è eø ρ E

−1. 565

L 1000 ρ = 2 : σf,cr * ( ) = 6148 * æç ö÷ è eø ρ E

−1. 565

L 1000 ρ = 1 : σf,cr * ( ) = 6977 * æç ö÷ è eø ρ E

−1. 565

L 1000 ρ = 0.5 : σf,cr * ( ) = 7886 * æç ö÷ è eø ρ E Revision 1 © AEROSPATIALE - 1998

−1. 565


page V1-2•4.25

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS c) in torsion critical torsion stress

Curve equations: curves valid in the elastic domain with: 5° < α < 11° L 1000 ρ = 8 : τt,cr * ( ) = 545 * æç ö÷ è eø ρ E

−1. 407

L 1000 ρ = 4 : τt,cr * ( ) = 598 * æç ö÷ è eø ρ E

−1.362

L 1000 ρ = 2 : τt,cr * ( ) = 695 * æç ö÷ è eø ρ E

−1.328

L 1000 ρ = 1 : τt,cr * ( ) = 764 * æç ö÷ è eø ρ E

−1. 285



page V1-2•4.26

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS d) under external pressure CRITICAL EXTERNAL PRESSURE

Z=

L2

ρ moy e

1 − ν 2e

Curve equation: 0 < Z < 60 : ky = 4 + 0.12241 * Z - 0.0006 * Z² 0.45 Z > 60 : ky = 1.456 * Z Critical external pressure: Pcr = η


π2 E e3 k y 12 (1 − ν 2e ) ρ moy L2


page V1-2•4.27

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •4.4-INTERACTION EQUATIONS (COMBINED LOADING) V1-2 •4.4.1-Circular and elliptical cylinders A - No external pressure a - Compression + bending + torsion + shear load The interaction equation of a circular cylinder subjected to axial compression combined with bending, torsion and shear due to a shear load is: Þ Interaction equation: Rc + R2ct + 3 R 3cs + R 3f = 1 where: Rc =

applied compression load (or applied compression stress) critical compression load (or critical compression stress)

Rct =

shear stress due to torsion critical torsion stress

Rcs =

shear stress due to the shear load critical shear stress

Rf =

bending moment applied (or bending stress) critical bending moment (or critical bending stress)

The intermediate interaction equations are derived from the general information above. Thus: 1 b - Longitudinal compression + pure bending: Rc + Rf = 1 ; M.S = -1 Rc + Rf c - Longitudinal compression + torsion:

2 Rc + R2ct = 1 ; M.S = -1 R c + R 2c + 4 R 2ct

d - Pure bending + shear load:

R3f + R3cs = 1 ; M.S =3

1 R + R 3cs 3 f

-1

e - Longitudinal compression + pure bending + torsion: 2 Rc + Rf + R2ct = 1 ; M.S = -1 R c + R f + ( R c + R f ) 2 + 4 R 2ct f - Longitudinal compression + pure bending + shear load: Rc + 3 R 3cs + R 3f = 1 refer to the charts on the following page. Revision 1 © AEROSPATIALE - 1998


page V1-2•4.28

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS Interaction curves under axial compression, bending, shear load



page V1-2•4.29

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS B - With internal pressure The internal pressure appreciably improves the cylinder's strength whatever the loading mode. Thus, the interaction formulas above remain valid provided that the new critical stresses, improved by the presence of an internal pressure, are taken into account. a) Compression + internal pressure: The increase in critical stress in compression due to internal pressure is given below: Increase in critical stress in compression due to internal pressure

∆σcc,0 r ηE e

100

An isostatic internal pressure p implies a longitudinal tension at the cylinder σt: σt =

pr 2e

A radial pressure implies nil longitudinal tension: σt = 0 The critical compression load is then: Pc,cr = 2 π r e (σcc,0 + ∆σcc,0 + σt) σcc,0, ∆σcc,0, σt are all positive values. The safety margin is then: M.S = Revision 1 © AEROSPATIALE - 1998

Pc,cr Papp

-1


page V1-2•4.30

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS b) Bending + internal pressure: The increase in critical stress in bending due to internal pressure is given below:

∆ct,0 ηE

r e

increase in critical stress in bending due to the internal pressure

An isostatic internal pressure p implies a longitudinal tension at the cylinder σt: σt =

pr 2e

A radial pressure implies a nil longitudinal tension: σt = 0 The critical bending moment is then: Mf,cr = π r² e (σcf,0 + ∆σcf,0 + σt) σcf,0, ∆σcf,0, σt are all positive values. The safety margin is then: M.S =


M f ,cr M app

-1


page V1-2•4.31

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS c) Torsion + internal pressure: The critical stress in torsion due to the internal pressure is given below: effect of the internal pressure on the critical torsion stress

The interaction equation is: R 2ct + Rp = 1 (Rp < 0, if p is an internal pressure) Rct =

applied torsion stress critical torsion stress

; Rp =

applied pressure critical pressure

d) Shear load + internal pressure The interaction equation is: R 2cs + Rp = 1 (Rp < 0, if p is an internal pressure) Rcs =

applied shear stress critical shear stress


; R p=

applied pressure critical pressure


page V1-2•4.32


V1-2 •4-CYLINDERS AND OTHER SHELLS

e r L

The thin shells discussed in this chapter are circular and elliptical cylinders, spherical caps and truncated cones.



page V1-2•4.1

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS A - SUMMARY OF CASES DEALT WITH CIRCULAR CYLINDERS (single load) compression, bending, shear, torsion, radial pressure, hydrostatic pressure

Compression

Bending

Shear

Torsion

Radial pressure

Hydrostatic pressure

ELLIPTICAL CYLINDERS (single load) compression, bending, shear, torsion

Compression

Bending

Shear

Torsion

SPHERICAL SHELLS (single load) external pressure

External pressure CONICAL SHELLS (single load) compression, bending, torsion, external pressure

Compression Revision 1 © AEROSPATIALE - 1998

Bending

Torsion


External pressure page V1-2•4.2

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS B - CIRCULAR CYLINDER ANALYSIS PROCEDURE Determine the following data: - cylinder geometry: L, e, r - cylinder loading, - properties of the cylinder material Identify the cylinder boundary conditions: free, hinged, fixed edges, etc.

Determine the buckling factors K (paragraph 4.3.1) Kc if compression Kf if bending Kt if torsion or shear load Ky if external radial pressure Select a plastic correction factor (paragraph 4.2)

Calculate the buckling elastic-plastic critical stresses (paragraphs 4.2 and 4.31): compression, bending, torsion, shear, radial pressure No

Is the cylinder subjected to combined loading?

Calculate Ri

Yes

with internal pressure? No Yes

Calculate the loading rate (paragraph 4.4.1.A) Rc if compression Rcs if shear Rct if torsion Rf if bending

Calculate the loading rate (paragraph 4.4.1.B) Rc if compression Rcs if shear Rct if torsion Rf if bending

Use the interaction equation Rc + Rct^2 + (R cs^3 + R f^3)^0.33 = 1 Revision 1 © AEROSPATIALE - 1998


page V1-2•4.3

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS C - ELLIPTICAL CYLINDER ANALYSIS PROCEDURE Determine the following data: - geometry of the cylinder: L, e, a, b - cylinder loading - properties of the cylinder material

Identify the cylinder boundary conditions: free, hinged, fixed edges, etc.

Determine the buckling factors K (paragraph 4.3.2) Kc if compression Kf if bending Kt if torsion or shear

Select a plastic correction factor (paragraph 4.2)

Calculate the buckling elastic-plastic critical stresses (paragraphs 4.2 and 4.3.2) in: compression, bending, torsion, shear Yes

No

Is the cylinder subject to combined loading?

with internal pressure?

Calculate Ri No

Yes Calculate the loading rates (paragraph 4.4.1.A) Rc if compression Rcs if shear Rct if torsion Rf if bending

Calculate the loading rates (paragraph 4.4.1.B) Rc if compression Rcs if shear Rct if torsion Rf if bending

use the interaction equation Rc + Rct^2 + (R cs^3 + R f^3) ^ 0.33 = 1 Revision 1 © AEROSPATIALE - 1998


page V1-2•4.4

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS D - SPHERICAL SHELL ANALYSIS PROCEDURE Determine the following data: - shell geometry: d, e, r - shell loading: external pressure - properties of the shell material Determine the buckling factor Kp (paragraph 4.3.3) Select a plastic correction factor (paragraph 4.2) Calculate the buckling elastic-plastic critical stress (paragraphs 4.2 and 4.33) under: external pressure

E - CONICAL SHELL ANALYSIS PROCEDURE Determine the following data: - shell geometry: L, e, r, R, α, ρ - shell loading - properties of the shell Determine the buckling factors K (paragraph 4.3.4) Kc if compression Kf if bending Kt if torsion Ky if external pressure Select a plastic correction factor (paragraph 4.2) Calculate the buckling elastic-plastic critical stresses (paragraphs 4.2 and 4.3.4) in: compression, bending, torsion, external pressure No

Is the cylinder subject to combined loading?

Calculate Ri

Yes

Calculate the loading rates (paragraph 4.4.2) Rc if compression Rf if bending Rct if torsion Ry if external pressure use the interaction equation (Rc + Rf + Rct) ^ 1.2 + R y ^ 1.2 = 1



page V1-2•4.5

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •4.1-BEHAVIOUR OF CIRCULAR CYLINDERS UNDER COMPRESSION V1-2 •4.1.1-Introduction Circular cylinders may have 4 different types of behaviour under the effect of compression: • Short cylinders The deflected shape after buckling is similar to that of column plates (cylinders with an infinite radius). The buckling waves are similar to those obtained in flat plates (half sine waves). The supporting conditions at ends have a considerable influence on the critical stress.

• Long cylinders The deflected shape after buckling is of a diamond-like appearance. The supporting conditions at ends have no affect on the critical stress.

• Transition length cylinders The transition length is in between the two extreme lengths defining the short cylinders and the long cylinders. The buckle pattern is an interaction of two shapes: This is a combination of the half sine waves and diamond shape.

• Very long cylinder A cylinder is very long when its column buckling stress is less than its local buckling stress.



page V1-2•4.6

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •4.1.2-Linear theory and test results Here we shall deal with the first 3 types of behaviour mentioned above. The energy causing the occurrence of an inward blister is much lower than that required to obtain an outward blister. With compressed cylinders, unlike what happens in flat plates, the occurrence of an inward blister entails the formation of a transverse compression membrane stress. This stress make the blister even more unstable. The linear stability theory (low deflection) gives satisfactory results when positive or nil membrane transverse stresses (tension) accompany folding. This is the case for flat plates, cylinders subjected to torsion or radial pressure, curved plates in shear or under side pressure. On the other hand, for curved plates and cylinders in compression or bending and for spheres or spherical caps subjected to an external radial pressure, the surface stresses can no longer be considered as linear functions of transverse displacements. The formulas and charts given in this document are derived from the linear theory readjusted following a great number of tests. A good correlation of tests and stresses calculated in this manner is obtained for short or long cylinders in compression or bending. On the other hand, the results are highly conservative in the transition range. In this latter case, the test results are located between the theoretical linear calculations and the semi-empirical calculations.



page V1-2•4.7

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •4.2-PLASTICITY CORRECTION FACTOR Even if the studied shell or cylinder is subjected to combined loads, the plasticity correction is applied to the buckling stresses under single loading. CIRCULAR AND ELLIPTICAL CYLINDERS Cylinder loading

Plasticity correction factor η=

Compression, Bending

Es E

η=

Shear, Torsion Radial or hydrostatic pressure

η=

Et Es Gs G

E s 1 − ν e2 æ E ö 0.25 + 0.75 t ÷ 2 ç E 1− ν è Es ø

SPHERICAL CONES AND TAPERED CONES Cone loading

Plasticity correction factor

η=

External radial pressure

Es E

Et Es

In the particular case of shear, the compression stress-strain curve of the material is also used. Proceed as follows: • Calculate the equivalent normal stress:σeq = τ 3 • Calculate Es and ν corresponding to σeq, G æ 1 − νe ö Es • s =ç ÷ G è 1 − ν2 ø E



page V1-2•4.8

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS V1-2 •4.3-BUCKLING FACTORS (SINGLE LOADING) V1-2 •4.3.1-Circular cylinders Six different loadings cause the occurrence of stress in the surface of a cylinder:

Compression

Bending

Shear

Torsion

Radial pressure

Hydrostatic pressure

The corresponding buckling stresses are given in the paragraphs below in function of the geometry of the cylinder and the boundary conditions.

V1-2 •4.3.1.1-Single compression The critical buckling stress under single compression is formulated as below: 2 eö π2 æ σcc,0 = η Kc E ç ÷ with Kc = kc è Lø 12 (1 − ν 2e ) Calculation of kc L2 The buckling factor depends on the parameter: Z = 1 − ν 2e re * short cylinders and transition range kc0: buckling factor of column plates 2 (0.581 x C x Z) fixed edges: kc0 = 4 kc = kc0 + k c0 hinged edges: kc0 = 1

* long cylinders kc = 1.16 x C x Z

Þ

e σcc,0 = η C E æç ö÷ è rø

à C factor The C factor used in the formulas below is: −0 .15 æ ö rö r æ C = 0.6 x ln ç 3.748 x ç ÷ ÷ ; C≤ 0.6 valid for: 8.52 < < 3000 è ø e e è ø r if < 8.52 take C = 0.6 e à Border between the transition range and long cylinders: k Z* = c0 and kc* = 2 kc0 0.58 C Revision 1 © AEROSPATIALE - 1998


page V1-2•4.9

SSM V1-2 • BUCKLING OF THIN PLATES AND SHELLS Buckling factor in compression kc as a function of Z loaded sides fixed



loaded sides simply supported

page V1-2•4.10


YES

X

NO


1 4

5 4

2 1 2

Volume 1

5


Purpose

Scope


Contents

Document Manager



Not applicable.

V1 - 1 V1 - 2 V1 - 3 V1 - 4 V1 - 5 V1 - 6 V1 - 7 V1 - 8 V1 - 9



Validation

Name: J. HUET



3page 1

Title - Annex

Reference documents


Abbreviations




Definitions


Highlights

Issue

Date

Pages modified

A

02/98

V1 - 1 à V1 - 3 V1 - 7 à V1 - 9

B

05/99

V1 - 7


V1 - 4

New chapter.

V1 - 1


C

11/99




New chapter.

MTS 004 Iss. C

3Ann. page



List of approval

Dept. code

Function

BTE/CC/CM

Name / First name


Key words

Calcul

Bibliography

Néant

Signature

CAZET G.

Distribution list

Dept. code

Function


BQP/TE

Archives Diderot

SIBADE Alain

BQP/TE


SIBADE Alain


BOUTET Fernand

BTE/SM/MG



MTS 004 Iss. C

page IG1

Static stressing manual

V1-3 •

STIFFENERS

A - CONTENTS Chapter V1.3 V1.3.1 V1.3.2 V1.3.3 V1.3.4 V1.3.5 V1.3.6 V1.3.7 V1.3.8

Title

Issue 1 1 1 1 1 1 1 1 1

STIFFENERS GENERAL DATA PREPARATION LOCAL BUCKLING WARPING FORCED CRIPPLING INTER-RIVET BUCKLING CRIPPLING OPTIMIZATION

Date 12 / 1996 12 / 1996 12 / 1996 12 / 1996 12 / 1996 12 / 1996 12 / 1996 12 / 1996 12 / 1996

Original issue Original issue Original issue Original issue Original issue Original issue Original issue Original issue Original issue Pages

V1.3 STIFFENERS A - Contents B - Symbols C - References

V1.3.1 V1.3.1 V1.3.3 V1.3.5

V1.3.1 GENERAL

V1.3.1/1

V1.3.2 DATA PREPARATION

V1.3.2/1

V1.3.3 LOCAL BUCKLING

V1.3.3/1

V1.3.3.1 Assumptions

V1.3.3/1

V1.3.3.2 Local buckling of the flange V1.3.3.2.1 Flat flange V1.3.3.2.1.1 Critical local buckling stress σflt V1.3.3.2.1.2 Local buckling factor V1.3.3.2.1.3 Plasticity correction factor V1.3.3.2.2 Curved flange V1.3.3.2.2.1 Critical local buckling stress σflt V1.3.3.2.2.2 Local buckling factor V1.3.3.2.2.3 Plasticity correction factor V1.3.3.2.3 Block diagram V1.3.3.2.4 Example

V1.3.3/2 V1.3.3/2 V1.3.3/3 V1.3.3/3 V1.3.3/4 V1.3.3/5 V1.3.3/5 V1.3.3/6 V1.3.3/6 V1.3.3/7 V1.3.3/8

V1.3.3.3 Local buckling of the web V1.3.3.3.1 Critical local buckling stress σfla V1.3.3.3.2 Local buckling factor kc V1.3.3.3.3 Plasticity correction factor V1.3.3.3.4 Block diagram V1.3.3.3.5 Example

V1.3.3/11 V1.3.3/12 V1.3.3/12 V1.3.3/13 V1.3.3/15 V1.3.3/16

Page V1-3 / 1 revision 1

© Copyright AEROSPATIALE 1998


V1-3 •

STIFFENERS Pages

V1.3.3.4 Sizing of a flanged edge V1.3.3.4.1 Lip V1.3.3.4.2 Bulb V1.3.3.4.3 Block diagram

V1.3.3/19 V1.3.3/19 V1.3.3/24 V1.3.3/25

V1.3.4 WARPING

V1.3.4/1

V1.3.4.1 Assumptions V1.3.4.2 Stiffness of the equivalent elastic medium V1.3.4.3 Warping stress (linear elasticity) ~ V1.3.4.4 Determination of m0 minimising σ dr ~ V1.3.4.5 Calculation of σ dr V1.3.4.6 Plasticity correction V1.3.4.7 Block diagram - summary V1.3.4.8 Example V1.3.5 FORCED CRIPPLING

V1.3.5/1

V1.3.5.1 Assumptions V1.3.5.2 Allowable stresses V1.3.5.3 Equivalent thickness of the stiffener skin side flange V1.3.5.4 Example V1.3.6 INTER-RIVET BUCKLING V1.3.6.1 Assumptions V1.3.6.2 Critical stress V1.3.6.3 Example

V1.3.4/1 V1.3.4/2 V1.3.4/3 V1.3.4/4 V1.3.4/5 V1.3.4/5 V1.3.4/7 V1.3.4/8

V1.3.5/1 V1.3.5/2 V1.3.5/5 V1.3.5/6 V1.3.6/1 V1.3.6/1 V1.3.6/1 V1.3.6/3

V1.3.7 CRIPPLING

V1.3.7/1

V1.3.8 OPTIMISATION

V1.3.8/1




V1-3 •

STIFFENERS

B - SYMBOLS A: b: C: e: Ec: ES: ET: f: h: I: k c: k: K: L: m: n: p: R: r: ν: νe: νp: η: β: λ: σdr: σcrip: σfla: σflf: σflt: σir: σrTDmin: σc0.2: τ: τcp: τflf: ~: σ

cross section area width buckling coefficient for a curved flange thickness Young's modulus in compression secant modulus tangent modulus deflection height Inertia buckling factor for a flat item diagonal tension factor end fixity factor length number of half wave lengths stress-strain shape factor (Ramberg and Osgood) inter-rivet pitch curvature radius radius Poisson's ratio in the elastic-plastic field Poisson's ratio in the elastic domain Poisson's ratio in the plastic field plasticity correction ratio elastic foundation modulus column slenderness critical flange warping stress critical crippling stress critical buckling stress of the web critical forced crippling stress critical buckling stress of flange critical inter-rivet buckling stress compression stress due to diagonal tension in the stiffener 0.2% yielding point (conventional allowable compression yield stress) shear stress in the panel critical folding stress of the panel critical forced crippling shear stress stress calculated before plasticity correction




V1-3 •

STIFFENERS

Meaning of indexes a: c: bt: p: r: s: t: tp: x, y, z:

"web" "compression" "flanged edge" "skin" "stiffener" "skin side flange" "flange" "pad" "reference axes"




V1-3 •

STIFFENERS

C - REFERENCES

REF. 1:

Theory of elastic stability

TIMOSHENKO

REF. 2:

Analysis and design of flight vehicle structures

BRUHN




V1-3 •

STIFFENERS





V1-3 •

STIFFENERS

V1-3 •1-GENERAL The purpose of this chapter is to make it possible to: 1 - calculate the various allowable stresses in compression proper to stiffeners (consequently, the following shall be considered as a complement to chapter V1-1 STIFFENED PANELS). 2 - optimise a stiffener in the design phase.

V1-3 •1.1-DEFINITIONS Stiffeners are members secured to the panel to which they are supposed to increase the stiffness. These members are laid in the normal dominating stress direction. For example, with a fuselage the stiffeners are called "stringers". If the panel and the stiffener are two distinct items, the stiffener is then called a "fastened stiffener" (it is fastened to the panel over its entire length with bolts, rivets or tack welds). If the panel/stiffener assembly is machined from a thick plate, the stiffener is then called an "integral stiffener". In all cases, the length of a stiffener is always much greater than its transverse dimensions. For local studies, these stiffeners are broken down into skin side flanges (secured to the skin), webs and flanges. CAP

FLANGED EDGE STIFFENER W E B

SKIN SIDE FLANGE

SKIN

Skin side flange: Part of the stiffener connected to the panel by riveting or welding, of which the centre plane is parallel to the centre plane of the panel. Cap: Free part of the stiffener of which the centre plane or the axis (curved cap stringer) is parallel to the centre plane of the panel. Web: Part of the stiffener connecting the flanges of which the centre plane is perpendicular or forms an angle with the centre plane of the panel. Flanged edge: Part of the stiffener connected to the cap and located opposite the web.

General - Page V1-3•1/1 revision 1



V1-3 •

STIFFENERS

V1-3 •1.2-BEHAVIOUR, FAILURE MODES Stiffeners are capable of handling a normal load Nx, a shear load Tz and a bending moment My.

panel

G My Nx Tz

Bending and normal loads result in stresses. The shear load causes shear stresses in the web but which may be considered as negligible (long beam assumption). The stiffener is mainly subjected to tension and compression stresses. * Tension: The static strength limit of the stiffener at UL is given by the properties. * Compression: This stiffener may be unstable under this type of load. Such instabilities are: - under a short wave length mode (local): Local buckling of the flange Local buckling of the web - under an average wave length mode: Stiffener warping - at the skin side flange and related to the fact that rivets are used: Forced crippling Inter-rivet buckling - related in general to the stiffener: Crippling resulting in stiffener failure. This phenomenon only occurs after local buckling. It is easy to understand the advantage of breaking down the stiffener into small flat or curved interconnected plates, in order to be able to study each of the phenomena separately.




V1-3 •

STIFFENERS

A stiffener is optimised if no local instability can occur in it before the conventional compression allowable compression yield stress. Its allowable stress is then σc0.2.

STIFFENER INSTABILITY

LOCAL BUCKLING OF THE CAP

WARPING

CRIPPLING (closed type cross section)

CAP

LOCAL BUCKLING OF THE WEB

FLANGED EDGE

W E B

STIFFENER

FORCED CRIPPLING

SKIN SIDE FLANGE

SKIN

INTER-RIVET BUCKLING




V1-3 •

STIFFENERS





V1-3 •

STIFFENERS

V1-3 •2-DATA PREPARATION V1-3 •2.1-MATERIAL DATA STABILITY CALCULATIONS: These calculations require the incorporation of the elastic-plastic behaviour of materials. To this end, the Ramberg and Osgood model is used (refer to V2-2 "Material behaviour"): æ σ ö σ ε= + 0.002 . ç ÷ Ec è σ c 0. 2 ø σ Es = ε 1 n 1− n = + ET Es Ec

n

Therefore, the following is needed for each type of material used: • the Young's modulus in compression, cE, • the conventional allowable compression yield stress σc0.2, • the Ramberg & Osgood coefficient, n.

V1-3 •2.2-GEOMETRICAL DATA As the data required to study the stiffener are specific to each phenomenon, they are detailed in the paragraphs involved.

Data preparation - Page V1-3•2/1 revision 1



V1-3 •

STIFFENERS


Data preparation - Page V1-3•2/1 revision 1



V1-3 •

STIFFENERS

V1-3 •3.4.2-Bulb bt

For a bulb flange there is: πD 2bt Abt = 4 2 πD 4bt πD 2bt æ D bt − e t ö Ibt = + ç ÷ ø 64 4 è 2

Dbt

et

bulb flange

with Ibt inertia of the bead in relation to the neutral line of the flange.

Assumption: the critical buckling stress of the bead is always greater than that of the flange. In this case, there is only one condition to verify: The bead must be at least one support for the flange. 4

3

2

æb ö æD ö æD ö æD ö This means, using the Eq. 1.3.3.4.1: ç bt ÷ − 1.6 ç bt ÷ − 0.374 ç bt ÷ ≥ 7.44 ç t ÷ ; è et ø è et ø è et ø è et ø Eq. 1.3.3.4.2

By representing this equation by the following curve (Fig. 1.3.3.4.2.1), it is possible to determine the minimum diameter of the bead Dbt as a function of dimensions et and bt of the flange: Minimum values of the Dbt/ et ratio as a function of the dimensions bt and et of the flange, so that a bead is a support for the flange

bulb flange

Fig. 1.3.3.4.2

Local buckling - Page V1-3•3/24 revision 1



V1-3 •

STIFFENERS

V1-3 •3.4.3-Block diagram FLANGED EDGE SIZING

E

CONDITIONS

I A 1 - 2.73 bt 3 − bt ≥ 5 b t et bt et

and

~ ≥σ ~ 2- σ fl bt fl flange

Type of flanged edge?

lip Abt = bbt ebt æ 1ö Ibt = ç ÷ ebt è 3ø

bead

bt

b 3bt

Abt =

bbt

et

Ibt =

if ebt and bbt are to be defined

bt

( e bt ) min. ( b bt )selected

æe ö 0.55ç t ÷ è bt ø

2

πD bt

2ö æ æ ö b t ç 1 + 1 + 6ç ÷ ÷ ç è e t ø ÷ø è

64

πD bt æ D bt − e t ö ç ÷ ø 4 è 2 2

+

bulb flanged edge

4

3

2

æ D bt ö æD ö æD ö æb ö ç ÷ − 1.6ç bt ÷ − 0.374ç bt ÷ = 7.44ç t ÷ è et ø è et ø è et ø è et ø refer to Fig. 1.3.3.4.2

bt0

e bt

2

E N

if ebt to be defined and bbt imposed

and bt = bt0 -

S

Condition 2 is always verified and condition 1 imposes at least:

æe ö = 3.05ç t ÷ è bt ø

b bt

S

2

refer to Fig. 1.3.3.4.1.1

with βt =

Dbt

et

4 4

ebt lip

( b bt ) min.

πD

bt

2 bt

it is therefore necessary that:

ì ùü é ï úï ê e bt et ù ê 5 ïé úï ≥ max í ê1 − 1 − 6.1β t ý ú;ê 3 bt0 bt0 û æ b t 0 ö úï ïë 3 æ bt0 ö ê 2.5 + 0.91β t ç ÷ − β t ç ÷ ú ï è et ø è e t ø úû ïþ êë î e bt ≤1 b t0

T I

refer to Fig. 1.3.3.4.1.2 if bbt to be defined and ebt imposed

by putting down αt =

e bt

æ 0.55 æ 5 b t çç ≥ ç3 + ç et è α t è α t et

b bt

A

it is necessary that:

et 2 ö ö ö÷ 0.55 æ 5 b t 25 b t ÷ ç − 0.163 ÷ + 3 − − 0.163 ÷ 2 2 2 2 αt et α t è α t et αt et ø ÷ø ø 2

25 b t

b ö æ ≤ ç 0.328α t ÷ t e et è t ø

b bt

refer to Fig. 1.3.3.4.1.3



L


V1-3 •

STIFFENERS





V1-3 •

STIFFENERS

V1-3 •3.3.3-Plasticity correction factor ~ is calculated with η = 1, The critical stress noted σ fla ~ ≥ 0.5 . σ ~ if: σ then σ fla c0.2 fla = η . σ fla, This calculation is iterative as η = ƒ (ES, ET, ν) = g(σfla). It converges when:

σ fla ~ →σ fla η

To be applied in the case when: σfla ≥ 0.5 . σc0.2 According to Chapter V.1.2 (paragraph 2.2) of this manual, there is: æ 1 − ν 2e ö E S - For a flange-type plate (no cap): η1 = ç ÷ è 1 − ν2 ø E c æ E ö - For a plate simply supported in compression: η3 = η1 ç 0.5 + 0.25 1 + 3 T ÷ ES ø è Eq. 1.3.3.3.3

with: Ec: ES: ET: ν e: ν:

Young's modulus in compression secant modulus tangent modulus Poisson’s ratio in the elastic domain Poisson’s ratio in the elastic-plastic domain

Remark: If the stiffener does not have a cap, then refer to the previous chapter (V1.3.3.2) for calculation of the local buckling of the web proceeding as for a flat flange without a flanged edge (side skin flange-type plate).




V1-3 •

STIFFENERS





V1-3 •

STIFFENERS

V1-3 •3.3.4-Block diagram

E

LOCAL BUCKLING OF THE WEB

CALCULATION OF kc

S

Is the flange a support for the web?

yes

It

2.73

3

haea

kc = 4

At

−

no

S

≥5

haea

Eq. 1.3.3.3.2.1 kc = 4 + 0.714

éæ êç 2.73 ëè

It 3

h a ea

−

At haea

ö ÷ ø

− 5

ù ú û

E

CALCULATION OF σfla ~ σ fla

= kc

æ ea ö ÷ 2 ç 12 (1 − ν e ) è h a ø Ec π

2

2

Eq. 1.3.3.3.1

N

PLASTICITY CORRECTION no

~ σ fla

≥ 0.5 . σc0.2

T

yes

~ if flange • σfla = η3 . σ fla ~ • σfla = η1 . σ fla if no flange

with:

æ

ET ö

è

ES ø

η3 = η1 ç 0.5 + 0.25 1 + 3

÷

and:

æ 1 − ν 2e ö E S ÷ è 1 − ν2 ø E c

η1 = ç

Eq. 1.3.3.3.3 end


I


A L


V1-3 •

STIFFENERS

V1-3 •3.3.5-Example Use the stiffener studied in this chapter: 16 4

30

2 3

2 3

25 35

The geometrical properties are: (refer to paragraph V1.3.1) cap thickness: cap width: web thickness: web height:

et = 4 mm bt = 15 mm ea = 2 mm ha = 26,5 mm

The characteristics of the materials are: (refer to paragraph V1.3.2) - stiffener: 7075 T73510

σr = 495 MPa σc0.2 = 420 Mpa νe = 0.33

Ec = 73800 MPa n = 19.66 νp = 0.5

V1-3 •3.3.5.1-Calculation of kc e t b 3t bt et 4 (15) 3 4 .15 0.91 − = 0.91 − = 56.8 ≥ 5 3 3 h a ea ha ea 26.5 (2 ) 26.5 . 2 The condition Eq.1.3.3.3.2.1.1 is verified and therefore the flange is a sufficient support for the web. kc = 4

V1-3 •3.3.5.2-Calculation of σfla 2

~ =k σ fla c

E c π2 æ ea ö 73800 ( π ) 2 ç ÷ =4 12 (1 − ν e2 ) è h a ø 12 (1 − (0.33) 2 )

æç 2 ö÷ è 26.5 ø


2

let:

~ = 1552 MPa σ fla



V1-3 •

STIFFENERS

V1-3 •3.3.5.3-Plasticity correction

~ σfla = η3 σ fla with: æ E ö η3 = η1 ç 0.5 + 0.25 1 + 3 T ÷ ES ø è

ν=

and

æ 1 − ν 2e ö E S η1 = ç ÷ è 1 − ν2 ø E c

æ ES E ö ν e + ç1 − S ÷ ν p Ec Ec ø è

Es =

ET =

1 1 0.002 æ σ fla ö + ç ÷ Ec σ c 0 .2 è σ c 0. 2 ø

n −1

1 1 0.002 n + Ec σ c 0. 2

æ σ fla ö ç ÷ è σ c 0. 2 ø

n −1

After numerical resolution, the following is found: σfla = 476 Mpa Remark: This critical local buckling stress of the web is greater than the 0.2% yielding point.




V1-3 •

STIFFENERS





V1-3 •

STIFFENERS

V1-3 •3.4-SIZING OF A FLANGED EDGE Flanged edges are of two types: lips and beads. bt

bt bbt

et

D

et

ebt bead

lip

The purpose of a flanged edge is to provide, at least, a single support for the flange. However, for this flanged edge to be efficient, it must not buckle before the flange. These two conditions are translated as follows: - The flanged edge must be a support for the flange: (Ref. 2, paragraph C7.9) 2.73

I bt A − bt ≥ 5 3 bt et bt et

Eq. 1.3.3.4.1

with: Abt: cross section area of the flanged edge Ibt: inertia of the flanged edge in relation to the neutral line of the flange - The flanged edge must not buckle before the flange: the critical local buckling stress of the flanged edge must be greater than that of the flange. σflbt ≥ σfltalon Eq. 1.3.3.4.2

V1-3 •3.4.1-Lip For a lip, there is: Abt = bbt ebt 1 Ibt = æç ö÷ e bt b 3bt è 3ø

bt

bending axis bbt

et ebt lip




V1-3 •

STIFFENERS

Being, for the first condition (Eq. 1.3.3.4.1): 3

æb ö b b 0.91 ç bt ÷ − bt − 5 t ≥ 0 et e bt è et ø

Eq. 1.3.3.4.1.1

For the second condition, the limits are set in the elastic domain as, if this condition is satisfied in this domain, it is all the more so satisfied in the plastic domain. In fact, if put down: - η1 and η3 the plasticity correction factors respectively of the flanged edge and the flange then: η1 ≥ η3 (refer to Chapter V.1.2 (paragraph 2.2)) of this manual. ~ the critical buckling stress calculated in the elastic domain, such as: σ = η σ ~ - σ fl fl fl then: ~ ~ ≥σ σ fl bt fl flange and Þ σ fl bt ≥ σ fl flange η1 ≥ η3 ~ ≥σ ~ The sufficient condition is therefore: σ fl bt fl flange which is formulated as follows: 2

kcbt

E c π 2 æ e bt ö EC π2 æ et ö ≥ k ç ÷ ç ÷ cflange 12 (1 − ν2e ) è b bt ø 12 (1 − ν2e ) è b t ø

2

Þ

b bt æ k cbt ö æ b t ö ÷ç ÷ ≤ç e bt çè k cflange ÷ø è e t ø

Eq. 1.3.3.4.1.2

with the kc factors given in the previous chapter: kc flange = 4: for the flange kc bt = 0.43: for the flanged edge The minimal ebt equation is deduced: æ et ö æ 4 ö æ et ö ebt = ç ÷ ç ÷ bbt let ebt = 3.05 ç ÷ bbt è 0.43 ø è b t ø è bt ø Using this equation in Eq.1.3.3.4.1.1 it is possible to obtain the minimum value of the ratio ( b bt ) min. e as a function of the ratio t of the flange, so that the lip is at least a single support bt bt for the flange: ( b bt ) min. 1 = bt 1.82

æ et ö ç ÷ è bt ø

2

æ ç1 + 1 + 18.2 æç 0.43 ö÷ ç è 4 ø è

æ bt ö ç ÷ è et ø

let:

2

ö ÷ ÷ ø

Eq. 1.3.3.4.1.3 æe ö ( b bt ) min. = 055 ç t ÷ bt è bt ø

2


2 æ ö ç 1 + 1 + 6 æç b t ö÷ ÷ ç è e t ø ÷ø è



V1-3 •

STIFFENERS

Having chosen a value for bbt so that (bbt)selected ≥ (bbt)min., it is necessary to determine the minimum width of the flanged edge so that it does not buckle before the flange. To this end, Eq. 1.3.3.4.1.2 is used:

(e bt ) min. ( b bt ) selected

æ 4 ö =ç ÷ è 0.43 ø

æ et ö ç ÷ let è bt ø

(e bt ) min. ( b bt ) selected

æe ö = 3.05 ç t ÷ Eq. 1.3.3.4.1.4 è bt ø

Finally, the straight flanged edge dimensions should verify that: bbt ≥ (bbt)min. ebt ≥ (ebt)min. with: Minimum dimensions ebt and bbt of a flanged edge as a function of the flange dimensions. 1 - bbt minimum as a function of bt and et. 2 - ebt minimum as a function of bt, et and bbt selected in 1.

(ebt)min.

(bbt)min.

bt

1

et

(bbt)selected

2

bt Fig.: 1.3.3.4.1.1




V1-3 •

STIFFENERS

Special cases: Case 1: If bbt is imposed then by putting down βt =

b bt e , such that bt = bt0 - bt and bt0 fixed: b t0 2

bt0

it is necessary: - according to Eq. 1.3.3.4.1.1: e bt 5 ≥ 3 b t0 æ b t0 ö 3 æ b t0 ö 2.5 + 0.91β t ç ÷ ÷ − βt ç è et ø è et ø

et

bbt

bt

- according to Eq. 1.3.3.4.1.2: 2

æ e bt ö æe ö e ç ÷ − 2 ç bt ÷ + 6.1 . β t t ≤ 0 b t0 è b t0 ø è b t0 ø e The second condition is only possible if: t ≥ 6.1 . βt, then in this case it is necessary: b t0

1 − 1 − 6.1β t

ebt

et e e ≤ bt ≤ 1 + 1 − 6.1β t t b t0 b t0 b t0

Furthermore, the case ebt ≥ bt0 means making the flange thicker, which is not the purpose of e this study. Therefore, work is limited to cases: ebt ≤ bt0 Þ bt ≤ 1; b t0 b e To conclude, if bbt is imposed with βt = bt and bt = bt0 - bt , then it is necessary: 2 b t0 ì é ùü ï ê úï et ù ê 5 ïé ú ï e bt max í ê1 − 1 − 6.1β t ú;ê 3 ú ý ≤ b ≤1 b t0 û æ ö æ ö b b t0 ïë ê 2.5 + 0.91β 3 ç t 0 ÷ − β ç t 0 ÷ ú ï t ï ï ê è et ø è e t ø úû þ ë î

Eq. 1.3.3.4.1.5

ebt must be selected in the following non-hatched domain: (case where β = 0.25). Determination of the thickness ebt of a flanged edge the height of the flanged edge bbt is imposed in relation to dimensions et and bt0 of the flange here b bt = 0.25 b t0

Fig.: 1.3.3.4.1.2




V1-3 •

Case 2: If ebt is imposed, then by putting down αt = æ ç 3 0.55 ç αt è

æ 5 bt ç çα e + è t t

STIFFENERS

e bt , it is necessary that: et

ö 25 b 2t 0.55 æ 5 b t ç ÷ 3 + − 0 . 163 − ÷ α 2t e 2t α t çè α t e t ø

ö ö÷ b bt æ b ö 25 b 2t ÷ ≤ ≤ ç 0.328α t t ÷ − 0 . 163 2 2 ÷ αt et et ø ø ÷ø e t è Eq. 1.3.3.4.1.6

For example, this is the case of a bent over edge. Then there is αt =

e bt = 1 and bbt has to be et

selected in the following non-hatched domain: Dimensions of the straight flanged edge as a function of the dimensions of the flange if they both have the same thickness et: α = 1

The flanged edge buckles before the flange

The flanged edge is not a support for the flange

Fig.: 1.3.3.4.1.3




V1-3 •

STIFFENERS

V1-3 •3-LOCAL BUCKLING V1-3 •3.1-ASSUMPTIONS For these calculations a stiffener angle section may be broken down into plates, the unloaded edges of which are either hinged or free. Then the rules given in Chapter V1-2 are used "Buckling of thin plates and shells". Local buckling concerns all parts of the stiffener which are not connected with the skin (the cap, the web, etc.). This is low wave length instability. The length of the unloaded edges is very long in relation to the width of the parts making up the angle section and therefore in relation to the half wave length of the pattern. The cap or the web must be considered as being infinitely long plates in single compression and therefore the final constant buckling factors must be used. The dimensions used for local buckling calculation or sizing of the flanged edge are as follows: bt bt Dbt bbt

et

et

ebt flat flange with a lip edge

straight flange with bead

rt

et

ea ea

ha

ha curved flange


web



V1-3 •

STIFFENERS

V1-3 •3.2-LOCAL BUCKLING OF THE FLANGE V1-3 •3.2.1-Flat flange

The boundary conditions at unloaded edges depend on two points: - Stiffness of the link between the flange and the web, - Whether or not there is a flanged edge and the amount of support provided by it. * Conservatively in all cases, the web-flange connection shall be considered as a single support. * The condition on the extreme edge of the flange depends on the presence or absence of a flanged edge: if there is a flanged edge, take into account the amount of support it provides in relation to the flange; otherwise, it is a free edge.

et

et bt

L

without a flanged edge


bt

L

with a flanged edge



V1-3 •

STIFFENERS

V1-3 •3.2.1.1-Critical local buckling stress σflt To calculate the critical local buckling stress of a flat flange, the results from Chapter V.1.2 (paragraph 2.3.1.1) of this manual (flat plate buckling) are used. In all cases (whether a flanged edge or not) the equation of the critical local buckling stress of the flat flange shall be: ηEc π2 σflt = kc 12 (1 − ν e2 )

æ et ö ç ÷ è bt ø

2

Eq. 1.3.3.2.1.1

with: bt: et: Ec: ν e: k c: η:

flange width flange thickness Young's modulus in compression Poisson ratio in the elastic domain local buckling factor (dépendant des C.L. et de la géométrie) plasticity correction factor

To calculate this critical stress, it is therefore necessary to determine the local buckling factor kc and the plasticity correction factor η, as a function of the boundary conditions and the geometry of the stiffener.

V1-3 •3.2.1.2-Local buckling factor - without a flanged edge:

kc = 0.43

(one edge supported, one free edge)

# If the flange has a flanged edge, then the factor kc of the flange depends on the amount of support this flanged edge provides. To ensure a simply supported condition, it has to satisfy the following conditions: (ref. 2, paragraph C7.9). 2.73

I bt A − bt ≥ 5 3 btet btet


Eq. 1.3.3.2.1.2.1



V1-3 •

STIFFENERS

kc is calculated as follows: - with a flanged edge such that: 2.73

I bt A − bt ≥ 5 then kc = kc∞ = 4 3 btet btet

- with a flanged edge such that: 2.73

éæ I bt A I A ö ù − bt < 5 then kc = 4 + 0.714êç 2.73 bt 3 − bt ÷ − 5ú 3 btet btet btet btet ø û ëè

Eq. 1.3.3.2.1.2.2 with: bt: flange width et: flange thickness Abt: cross section area of the flanged edge Ibt: inertia of the flanged edge in relation to the neutral line of the flange (refer to paragraph 3.4 for the equations of A1 and I1 in the case of a lip or a bulb (bead))

V1-3 •3.2.1.3-Plasticity correction factor ~ is calculated with η=1, The critical stress noted σ flt ~ ≥ 0.5 . σ ~ if: σ then σ flt c0.2 flt = η . σ flt, This calculation is iterative as η = f (ES, ET, ν) = g(σflt). It converges when:

σ flt ~ →σ flt η

According to Chapter V.1.2 (paragraph 2.2) of this manual, there is: æ 1 − ν 2e ö E S - For a flange without a flanged edge: η1 = ç ÷ è 1 − ν2 ø E c æ E ö - For a flange with a flanged edge: η3 = η1 ç 0.5 + 0.25 1 + 3 T ÷ ES ø è

Eq. 1.3.3.2.1.3 with: ES: ET: Ec: ν e: ν:

secant modulus tangent modulus Young's modulus in compression Poisson’s ratio in the elastic domain Poisson’s ratio in the elastic-plastic domain

Remark: η3 < η1 in case of doubt, choose the conservative solution.




V1-3 •

STIFFENERS

V1-3 •3.2.2-Curved flange

For local buckling calculation of a curved flange, the stiffener flange is compared to an infinitely long cylinder under single compression (refer to paragraph V1.2.4 of this manual: "Long cylinder buckling").

rt

et

V1-3 •3.2.2.1-Critical local buckling stress σflt To calculate the critical local buckling stress of the curved flange, the results from Chapter V.1.2 (paragraph 4.4.1) of this manual (long cylinder buckling) are used. The equation for the critical local buckling stress of the flange shall be:

æe ö σflt = η CEc ç t ÷ è rt ø

Eq. 1.3.3.2.2.1

with: rt: et: Ec: C: η:

flange radius flange thickness Young's modulus in compression local buckling coefficient plasticity correction factors




V1-3 •

STIFFENERS

To calculate this critical stress it is therefore necessary to determine the local buckling coefficient C and the plasticity correction factor η.

V1-3 •3.2.2.2-Local buckling factor According to Chapter V.1.2 (paragraph 4.4.1) of this manual, considering the case of long cylinders: −0 .15 æ ö æ rt ö rt ç ÷ - For: 8.52 ≤ ≤ 3000: C = 0.6 * In ç 3.748 ç ÷ ÷ e et è ø t è ø

- For:

rt ≤ 8.52: C = 0.6 et

Eq. 1.3.3.2.2.2

V1-3 •3.2.2.3-Plasticity correction factor ~ calculated with η = 1, Noting the critical stress σ flt ~ ~ ≥ 0.5 . σ then σ if: σ flt c0.2 flt = η . σ flt, This calculation is iterative as η = f (ES, ET, ν) = g (σflt). It converges when:

σ flt ~ →σ flt η

According to Chapter V.1.2 (paragraph 4.3) of this manual, there is: E ET - For a long cylinder in compression: η = S Ec ES Eq. 1.3.3.2.2.3

with: Ec: Young's modulus in compression ES: secant modulus ET: tangent modulus




V1-3 •

STIFFENERS

V1-3 •3.2.3-Block diagram LOCAL BUCKLING OF THE FLANGE curved flange

CALCULATION OF kc

CALCULATION OF C

Flanged edge?

yes

yes

E

flat flange

flanged edge = hinged edge? I bt

2.73

A bt

−

3

btet

rt

no

≤ 8.52

rt

8.52 ≤

et

≤ 3000

et

no

≥5

S

btet

Eq. 1.3.3.2.2.2

Eq. 1.3.3.2.1.2.1 kc = 4 kc = 4 + 0.714

éæ êç 2.73 ëè

I bt 3

A bt

−

btet

btet

ö ÷ ø

− 5

ù ú û

kc = 0.43

kc = 0.6

Ec π

~ =k σ flt c

2

æe ö ç ÷ ) èb ø

12 (1 − ν e

−0 . 15

t

ö ÷ ÷ ø

E

CALCULATION OF σflt

2

t

2

æ ærö C = 0.6 • Inç 3. 748 ç ÷ ç èe ø è t

CALCULATION OF σflt Eq. 1.3.3.2.1

S

Eq. 1.3.3.2.2.1

~ = CE σ flt c

æe ö ç ÷ èrø

N

t

t

t

PLASTICITY CORRECTION ~ ≥ 0.5 . σ ? σ flt c0.2

no

flat flange

curved flange

flanged edge?

yes

no

~ σflt = η3 . σ flt

~ σflt = η1 . σ flt

with:

with:

æ

η3 = η1 ç 0.5 + 0.25 1 + 3

è

ET ö

÷ ES ø

and:

æ1 − ö ES ÷ è1 − ν ø Ec

η1 = ç

T

yes

ν 2e 2

~ σflt = η . σ flt

I

with:

æ1 − ö ES ÷ è1 − ν ø Ec

η1 = ç

Eq. 1.3.3.2.1.3

ν 2e 2

η=

ES

ET

Ec

ES

Eq. 1.3.3.2.2.3

A L

Eq. 1.3.3.2.1.3 end




V1-3 •

STIFFENERS

V1-3 •3.2.4-Example Use the stiffener studied in Chapter V1-1: 16 4

30

2 3

2 3

25 35

The geometrical characteristics are: (refer to paragraph V1.3.1) flange thickness: et = 4 mm flange width: bt = 15 mm The materials properties are: (refer to paragraph V1.3.2) - stiffener: 7075 T73510 σr = 495 MPa Ec = 73800 MPa σc0.2 = 420 MPa n = 13.83 νe = 0.33 νp = 0.5

V1-3 •3.2.4.1-Calculation of kc No flanged edge, therefore:

kc = 0.43

~ (linear elastic material) V1-3 •3.2.4.2-Calculation of σ flf 2 2 Ec π2 æ et ö 73800 ( π ) æ 4ö = 0.43 ç ÷ ç ÷ 2 12 (1 − ν 2e ) è b t ø 12 1 − ( 0.33) è 15 ø 2

~ =k σ flt c

(


)

let:

~ = 2082 MPa σ flt



V1-3 •

STIFFENERS

V1-3 •3.2.4.3-Plasticity correction ~ σflt = η1 σ flt with: æ 1 − ν 2e ö E S η1 = ç ÷ è 1 − ν2 ø E c

æ E ö ES νe + ç1 − S ÷ νp Ec ø Ec è 1 Es = n −1 1 0.002 æ σ flt ö + ç ÷ Ec σ c 0 .2 è σ c 0 . 2 ø ν=

After numerical resolution there is:

σflt = 504 MPa Remark: This critical local buckling stress of the flange is greater than the 0.2% yielding point.




V1-3 •

STIFFENERS





V1-3 •

STIFFENERS

V1-3 •3.3-LOCAL BUCKLING OF THE WEB

The boundary conditions on the unloaded edges depend on the stiffness of the connections between the web and the cap, between the web and the skin side flange. * Conservatively and in all cases, the web-skin side flange connection is considered as a single support. * The web-cap connection depends on the amount of support the flange provides in relation to the web.

ha ea L

Remark: If the stiffener is not provided with a cap, then it is necessary to refer to the previous chapter (V1.3.3.2) to calculate the local buckling of the web, proceeding as for a flat flange without a flanged edge (skin side flange-type plate).




V1-3 •

STIFFENERS

V1-3 •3.3.1-Critical local buckling stress σfla To calculate the critical local buckling stress of the web, the results from Chapter V.1.2 (paragraph 2.3.1.1) of this manual (flat plate buckling) are used. The equation of the critical local buckling stress of the web is: η Ec π2 æ ea ö σfla = kc ç ÷ 12 (1 − ν 2e ) è h a ø

2

Eq. 1.3.3.3.1

with: ha: ea: Ec: ν e: k c: η:

loaded edge length (web height) web thickness Young's modulus in compression Poisson’s ratio in the elastic domain local buckling factor plasticity correction factor

Therefore, to calculate this critical stress, the local buckling factor kc and the plasticity correction factor η, have to be determined in function of the boundary conditions and the geometry of the stiffener.

V1-3 •3.3.2-Local buckling factor kc For the flange to truly support the web, it has to satisfy the condition given in Eq. 1.3.3.2.1.2.1 (refer to paragraph C7.9): 2.73

It At − ≥5 3 h a ea haea

Eq. 1.3.3.3.2.1

At: cross section area of the flange It: inertia of the flange in relation to the neutral line of the web If the flange satisfies this condition then, according to Chapter V.1.2 (paragraph 2.3.1.1) in this manual, the web is considered as an infinitely long plate simply supported on unloaded edges: I At if 2.73 t 3 − ≥5 kc = kc∞ = 4 h a ea haea ù éæ I At ö kc = 4 + 0.714êç 2.73 t 3 − ÷ − 5ú h a ea haea ø û ëè

if

2.73

It At − <5 3 h a ea haea

Eq. 1.3.3.3.2.2




V1-3 •

STIFFENERS

V1-3 •8-OPTIMISATION Critical stresses in a correctly sized stiffener are very close one to the other and are greater than the conventional allowable yield stress of the material.

V1-3 •8.1-SKIN SIDE FLANGE The fastener pitch must be such that the inter-rivet buckling stress is high enough. Ideally, this stress must be greater than or equal to the conventional allowable compression yield stress of the stiffener. This gives: • countersunk head bolts, rivets:

p ≤ e

• flat head bolts, tack welding:

p ≤ e

1.38 σ c 0 .2 + 0.002 n Ec 1.70 σ c 0 .2 + 0.002 n Ec

The maximum fastener pitch shall depend on the thickness of the skin under the stiffeners. Then, care shall be taken to ensure that the thickness of the stiffener skin side flange is sufficient, and that the pitch selected is compatible with the diameter of the fasteners: 3≤

p ≤6 D

p = 4.5 is a frequently used value ("sealed pitch") D Example: p ≤ 10.2 (E = 70300 MPa, σc0.2 = 270 MPa, n = 7.05) e If the skin thickness is 3 mm, the following is necessary: p ≤ 30.6 mm

• Skin made of 2024 PLT3, riveted stiffeners:

p ≤ 7.5 (E = 73800 MPa, σc0.2 = 420 MPa, n = 13.83) e If the thickness of the base is 3 mm, the following is necessary: p ≤ 22.5 mm However, if p=30 mm, then it is necessary to have esr ≥ 4 mm

• Stiffener made of 7075 T73510

In the first case, the following fasteners shall be used: In the second case, the following fasteners:

Optimisation - Page V1-3•8/1 revision 1

p = 4.7), D p φ 6.35 ( = 4.7). D

φ 4.8 (



V1-3 •

STIFFENERS

The skin side flange width shall be determined using the diameter of the selected fasteners (minimum edge distance = 2D, minimum distance between the centreline of the fasteners and the web).

2D mini D If the skin is subjected to diagonal tension, the pad thickness under the stiffener shall ideally e be 1.5 times the thickness of the adjacent pockets: t = 1.5 e

et

e

V1-3 •8.2-WEB Ideally, folding should not occur before σc0.2. This is the case if the ratio

h satisfies the e

following conditions: • Plate with a free unloaded edge:

h ≤ e

• Plate with two hinged unloaded edges:

h ≤ e

0.66 σ c 0.2 + 0.002 Ec 1.75 σ c 0.2 + 0.002 Ec

Example: Stiffener made of 7075 T73510 (E = 73800 MPa, σc0.2 = 420 MPa, n = 13.83) with flange:

h ≤ 20 e

Generally, the height of the web is limited for space allocation reasons.




V1-3 •

STIFFENERS

V1-3 •8.3-FLANGE The local buckling condition is governed by the rule defined for the web. Ideally, it would also be better if the warping stress was greater (or equal) to the allowable yield stress of the material. If the stiffener is provided with a single flange (Z stiffener), this last condition is practically impossible to satisfy.

It is necessary, at least, for the flange to be a support for the web.

V1-3 •8.4-CRIPPLING Imposes a supplementary condition on the skin side flange: refer to paragraph 8.2 (plate with one unloaded free edge). Example: Stiffener made of 7075 T73510:

bs ≤ 7.5. This results in a skin side flange width equal to the es

fastener pitch.




V1-3 •

STIFFENERS





V1-3 •

STIFFENERS

V1-3 •5-FORCED CRIPPLING The forced crippling phenomenon may occur when the panels carry loads in diagonal tension (refer to paragraphs V1-1.5 and V1-1.6: stiffened panels in single nominal shear).

V1-3 •5.1-ASSUMPTIONS Waves caused by folding of panel bays induce local deflection in the skin side flange of the stiffener which increases when the thickness of the flange is thinner than that of the panel. The free edge (AA) of the flange tends to follow the waves of the skin whereas the edge it makes with the web (BB) remains straight.

Fig. 1.3.5.1.1

Forced crippling - Page V1-3•5/1 revision 1



V1-3 •

A

STIFFENERS panel

A

A.A

stiffener Fig. 1.3.5.1.2

Then the local bending stresses amplified by the compression are added to the compression stresses due to the diagonal tension.

V1-3 •5.2-ALLOWABLE STRESSES Forced crippling occurs when the maximum compression in the stiffener reaches a value defined by a empirical equation integrating the plasticity of the material:

E S S E N T I A L

STRAIGHT STIFFENER (e.g.: Stringer...R(Curvature radius) > 3800 mm) σflf = - 0.051 .

σ c 0. 2 r σ c 0.2 r + 0.002 E cr

. k 2/3

æ e' E. ö . çç sr cr ÷÷ è e p . E cp ø

1/ 3

Eq. 1.3.5.2.1 (Ref. 2, Fig. C11.38)

with: e'sr: Ecr: ep: Ecp: k: σc0.2:

equivalent thickness of the stiffener skin side flange compression of the stiffener Young's modulus panel thickness compression of the panel Young's modulus diagonal tension factor (paragraph V-1.1.5.4) conventional allowable compressive yield stress of the stiffener




V1-3 •

STIFFENERS

CURVED STIFFENER (e.g.: Frame...500 mm ≤ r (curvature radius) ≤ 3800 mm) σflf = - (0.511r + 3095.3) . 10-5 .

σ c 0. 2 r

æ e' E. ö . k 2 / 3 . çç sr cr ÷÷ è e p . E cp ø

1/ 3

σ c 0.2 r + 0.002 E cr Eq. 1.3.5.2.2 (Ref. 2, Fig. C11.38)

with: e'sr: Ecr: ep: Ecp: k: σc0.2: r:

equivalent thickness of the stiffener skin side flange compression Young's modulus of the stiffener panel thickness compression Young's modulus of the panel diagonal tension factor (§ V-1.1.5.4) conventional allowable compression yield stress of the stiffener stiffener curvature radius

The forced crippling calculation is to be integrated in the general iteration of the calculation of a stiffened panel when the critical stiffener stress calculation step is reached. At each iteration, k, according to the applied stress, is set. Like this it is possible to calculate the compression stress value applied to stiffener σ TD r min as well as the allowable stress value. The calculation converges when: σflf ≈ σ TD r min The calculated value of σflf must be clipped at σc0.2. In other words, forced crippling occurs when the shear stress in the panel reaches a critical value: τ = τflf To determine the load τ = τflf causing forced crippling of the stiffener skin side flange, it is necessary to increment τ (single shear stress in the panel) until

σ TD = σflf rmin



E S S E N T I A L


V1-3 •

STIFFENERS

The following method is used to estimate τflf with sufficient accuracy: RF(0) =

E S

σ flf ( 0 ) σ

τ (0) τ cp

It is possible to estimate RS(1) (iteration 1) using the following equation RS(1) = RF(0) (RS(0) - 1) + 1

Eq. 1.3.5.2.3

Knowing RS(1) and τcp, it is possible to deduce τ and therefore τflf: τflf = RS(1) . τcp

S E N

RS(0) =

TD r min( 0 )

Then to iterate until RF(i) ≈ 1, putting down:

Eq. 1.3.5.2.4 τflf(i) = RS(i) . τcp R ϕ (i ) − 1 k(i) = Sϕ R S (i) + 1

(*)

σ TD k(i ). R S (i ) r min ( i ) = TD k(0). R S (0) σ r min (0)

T I A L

with:

σflf(i): σrTDmin(i): τ(i): τcp: τflf:

Critical forced crippling stress at iteration (i) Compression stress due to diagonal tension in the stiffener at iteration (i) Shear stress in the panel at iteration (i) Critical panel folding stress Critical forced crippling shear stress

(*) Reminder: According to paragraph V1-1.6.4, if (RS ≥ 1) Þ (k(i) =

R ϕS (i ) − 1 ) R Sϕ (i ) + 1

with æç 0.5 < B < 1ö÷ Þ æ ϕ = 0.4343 + 260.6 e p A ö ç ÷ è è A ø rB ø æç 0.5 < A < 1ö÷ Þ æ ϕ = 0.4343 + 260.6 e p B ö ç ÷ è è B ø rA ø B ìæ öü ïïçè 0.2 < A < 0.5÷ø ïï æ ep ö Þ ç ϕ = 0.4343 + 521.2 ÷ í ý B rø ïæç 0.2 < < 0.5ö÷ ï è ø ïîè ï A þ where: A: transverse member pitch B: stiffener pitch r: panel curvature radius ep: panel pocket thickness




V1-3 •

STIFFENERS

V1-3 •5.3-EQUIVALENT THICKNESS OF THE STIFFENER SKIN SIDE FLANGE The presence of a panel over thickness under the stiffener (PAD) delays the occurrence of forced crippling. e' sr, the effective thickness of the skin side flange, is then greater than esr (Eq. 1-3.5.2.1 or 2):

ep

etp

esr

bsr

btp Fig. 1.3.5.3.1

Equivalent thickness of the skin side flange: 2 e' sr = e sr +

Ep Er

(e

tp

− ep

)

2

limited to

esr'≤ 1.35 . esr

Eq. 1.3.5.3.1

• if etp > 1.5 ep, take etp = 1.5 . ep in the calculation of e'sr with:

etp: esr: ep :

pad thickness stiffener skin side flange thickness panel pocket thickness



E S S E N T I A L


V1-3 •

STIFFENERS

V1-3 •5.4-EXAMPLE Use the stiffener studied in Chapter V1-1 considering the case when it is straight, then the case in which it is curved: 16 4

30

2 3

2 3

25 35

Stiffener material:

7075 T73510

σr = 495 MPa

Ecr = 73800 MPa

σc0.2 = 420 MPa stiffener skin side flange thickness: esr = 3 mm skin side flange width: bsr = 25 mm Panel material:

2024 PLT351

σr = 440 Mpa

Ecp = 70300 MPa

σc0.2 = 270 Mpa skin thickness: pad thickness: pad width: frames pitch:

ep = 2 mm etp = 3 mm btp = 35 mm L = 530 mm

V1-3 •5.4.1-Straight stiffener According to the example in paragraph V1-1.5.12, there is: k = 0.24 τ = 160 MPa τcp = 52 MPa σ TD r min = - 55 MPa

V1-3 •5.4.1.1-Equivalent thickness e' sr =

application: e' sr =

32 +

70300 (3 − 2 ) 2 73800

e sr2 +

Ep Er

(e

tp

− ep

)

2

e' sr = 3.15 < 1.35 . e sr = 4.05 e'sr = 3.15




V1-3 •

STIFFENERS

V1-3 •5.4.1.2-Normal critical stress In the case of a straight stiffener, the critical stress is formulated: σflf = - 0.051 .

σ c 0 .2 r

.k σ c 0.2 r + 0.002 E cr with k = 0.24 (refer to paragraph V1-1.5.12), therefore:

σflf = - 0.051 .

with σ

2/3

æ e 'E. cr ö ÷÷ . çç sr è e p . E cp ø

1/ 3

æ 315 . . 73800 ö 420 . 0.24 2 / 3 . ç ÷ è 2 . 70300 ø 420 + 0.002 73800

1/ 3

σflf = - 111 MPa TD r

min (0) = - 55.3 MPa (refer to paragraph V1-1.5.12), therefore: σ ( 0) − 111 RF(0) = TD flf = σ r min(0 ) − 55 RF(0) = 2.017

V1-3 •5.4.1.3-Critical shear stress with τ(0) = 160 MPa and τcp = 52 MPa (refer to paragraph V1-1.5.12), therefore: τ ( 0 ) 160 Rs(0) = = τ cp 52 Rs(0) = 3.0769

Calculation of RS(1): RS(1) = RF(0) (RS(0) - 1) + 1 RS(1) = 2.017 (3.0769 - 1) + 1 RS(1) = 5.189

Calculation of the new panel loading τ: τ(1) = RS(1) . τcp = 5.189 . 52

τ(1) = 269 MPa This new shear value in the panel will give a new loading factor k and therefore a new stiffener stress. By continuing the calculation loop in paragraph V-1-1.5.12, there is: TD R 0. 4343 − 1 k (1) R s (1) σ r min (1) S(1) k(1) = 0. 4343 Þ σ TD = 0.343; ≈ = 2.41 r min (1) = - 133 MPa ( 0 ) R +1 k (0 ) R s (0 ) σ TD min r S(1) RF(1) =

σ flf (1) 141 = σ TD 133 r min (1)


and

Þ σflf (1) = - 141 MPa Þ RF(1) = 1.062



V1-3 •

STIFFENERS

giving the new Rs: RS(2) = RF(1) (RS(1) - 1) + 1 RS(2) = 1.062 (5.1499 - 1) + 1 RS(2) = 5.45 Þ τ(2) = 283 MPa

This new shear value in the panel will give a new loading factor k and therefore a new stiffener stress. By continuing the calculation loop of paragraph V-1-1.5.12, there is: TD k(2) R s (2) σ r min (2) k(2) = 0.352; ≈ TD = 2.6 k(0) R s (0) σ r min (0)

Þ σ TD r min (2) = - 143 MPa

and Þ σflf (2) = - 143 MPa and therefore a new RF: RF(2) = 1.001

An RF very close to 1 is obtained after two iterations. The calculation is stopped here and therefore:

τflf = 283 MPa




V1-3 •

STIFFENERS

V1-3 •5.4.2-Curved stiffener According to the example in paragraph V1-1.6.13, there is: r = 2820 mm k = 0.393 τ = 160 MPa τcp = 57 MPa σ TD r min = - 76 MPa

V1-3 •5.4.2.1-Equivalent thickness e sr2 +

e' sr =

application: e' sr =

32 +

70300 (3 − 2 ) 2 73800

Ep Er

(e

tp

− ep

)

2

e' sr = 3.15 < 1.35 . e sr = 4.05 e'sr = 3.15

V1-3 •4.5.1.2-Normal critical stress With a curved stiffener, the critical stress is formulated: σflf = - (0.511 . r + 3095.3) . 10 .

σ c 0 .2 r

-5

σ c 0. 2 r + 0.002 E csr

.k

2/3

æ e' E. ö . çç sr cr ÷÷ è e p . E cp ø

1/ 3

with k = 0.393 (refer to paragraph V1-1.6.13), therefore: 1/ 3

æ 315 . . 73800 ö 420 . 0.393 2 / 3 . ç σflf = - (0.511 . 2820 + 3095.3) . 10 . ÷ è 2 . 70300 ø 420 + 0.002 73800 σflf = - 137 MPa TD with σ r min (0) = - 76 MPa (refer to paragraph V1-1.6.13), therefore: σ ( 0) − 137 RF(0) = TD flf = σ r min(0 ) − 76 RF(0) = 1.804 -5

V1-3 •5.4.1.3-Critical shear stress with τ(0) = 160 MPa and τcp = 57 MPa (refer to paragraph V1-1.6.13), therefore: τ ( 0 ) 160 Rs(0) = = 57 τ cp RS(0) = 2.807




V1-3 •

STIFFENERS

Calculation of RS(1): RS(1) = RF(0) (RS(0) - 1) + 1 RS(1) = 1.804 (2.807 - 1) + 1 RS(1) = 4.26

Calculation of the new panel loading τ: τ(1) = RS(1) . τcp = 4.26 . 57

τ(1) = 242 MPa

This new shear value in the panel will give us a new loading factor k and therefore a new stiffener stress. By continuing the calculation iteration of paragraph V-1-1.6.13, there is: TD R 0.8039 − 1 k (1) R s (1) σ r min (1) S(1) k(1) = 0.8039 Þ σ TD = 0.524; ≈ TD = 2.023 r min (1) = - 154 MPa R +1 k (0) R s (0) σ r min (0 ) S(1) RF(1) =

Þ σflf (1) = - 166 MPa

and

σ flf (1) 166.97 = σ TD 154.59 r min (1)

Þ RF(1) = 1.08

giving the new RS: RS(2) = RF(1) (RS(1) - 1) + 1 RS(2) = 1.08 (4.26 - 1) + 1 RS(2) = 4.52 Þ τ(2) = 257 MPa

This new shear value in the panel will gives us a new loading factor k and therefore a new stiffener stress. By continuing the calculation iteration of paragraph V-1-1.5.12, there is: TD k(2) R s (2) σ r min (2) k(2) = 0.541; ≈ TD = 2.22 k(0) R s (0) σ r min (0)

Þ σ TD r min (2) = - 169 MPa

and Þ σflf (2) = - 170 MPa and therefore a new RF: RF(2) = 1.007

An RF very close to 1 is obtained after two iterations. The calculation is stopped here and therefore:

τflf = 257 MPa




V1-3 •

STIFFENERS

V1-3 •6-INTER-RIVET BUCKLING The inter-rivet buckling phenomenon occurs when there is local buckling in the panel (or the stiffener skin side flange) between two consecutive rivets. It may occur if the panel is loaded in compression. To prevent this phenomenon from occurring, the compression stress in the panel (or in the stiffener skin side flange) is limited to a value that will be defined here after.

V1-3 •6.1-ASSUMPTIONS A 2b wide sheet strip is assimilated to a column plate (refer to V1-2).

A.A panel ep

A es

b b p

stiffener base

A

V1-3 •6.2-CRITICAL STRESS The critical inter-rivet buckling stress is given by:

π2 .Ec æ e ö σir = η5 ç ÷ 12 è Kpø

2

Inter-rivet buckling - Page V1-3•6/1 revision 1

Eq. 1-3.6.2.1



V1-3 •

STIFFENERS

with: Ec: η5: K: p: e:

the compression modulus of the material of the item involved E plasticity correction factor (column plate) η5 = t (refer to paragraph V1-2.2.2) Ec end fixity coefficient, depending on the type of fastener used inter-rivet pitch thickness of the item involved (skin or flange)

Tests have shown that the boundary conditions at the ends of the column depend on the type of fasteners used. The ultimate conditions that will be encountered will be located between the dual-hinged conditions (K = 1) and the dual-fixed conditions (K = 0.5). Countersunk head bolt (or rivet) K = 0.66:

E S S E N T I A L

æ eö σir = 1.89 . η5 . Ec ç ÷ è pø

2

Eq. 1-3.6.2.2

Tack welding or flat head bolt K = 0.54:

æ eö σir = 2.82 . η5 . Ec ç ÷ è pø

2

Eq. 1-3.6.2.3

with: Ec: p: e: η5:

compression modulus of the material of the item involved inter-rivet pitch panel or stiffener skin side flange thickness E plasticity correction factor η5 = t (refer to paragraph V1-2.2.2) Ec

During the design phase, work shall be organised to have σir ≥ σc0.2 so as to use only σc0.2 as the allowable compression value.




V1-3 •

STIFFENERS

V1-3 •6.3-EXAMPLE Skin inter-rivet buckling: The stiffener skin side flange and the panel are joined by countersunk head bolts, pitch 25 mm. ep = 3 mm countersunk head bolt => K = 0.66 p = 25 mm Ecp = 70300 MPa np = 7.05 Critical stress: Eq. 3.3.6.2σirp = 1.89 . η5 . Ecp 3 σirp = η5 . 1.89 . 70300 æç ö÷ è 25 ø

2

æ ep ö ç ÷ è pø

2

σirp = η5 . 1911 MPa

Plasticity correction: η5 =

Et 1 with Et = ( np −1) E cp 0.002 . n p æ σ irp ö 1 .ç + ÷ E cp σ c 0 .2 è σ c 0. 2 ø

giving: σirp =

1911 0.002 . E cp . n p æ σ irp ö .ç 1+ ÷ σ c 0. 2 è σ c 0. 2 ø

( np −1)

or: σirp =

1911 0.002 . 70300 . 7.05 æ σ irp ö 1+ .ç ÷ è 270 ø 270

6 . 05

after numerical resolution, the following is obtained for the panel


σirp = 289 MPa



V1-3 •

STIFFENERS

Stiffener inter-rivet buckling: er = 3 mm countersunk head bolt => K = 0.66 p = 25 mm Ecr = 73800 Mpa nr = 13.83 Critical stress: æe ö Eq. 1.3.6.2σirr = 1.89 . η5 . Ecr ç r ÷ è pø 3 σirr = η5 . 1.89 . 73800 æç ö÷ è 25 ø

2

2

σirr = η5 . 2008 MPa Plasticity correction: η5 =

Et 1 with Et = ( nr −1) E cr 0.002 . n r æ σ irr ö 1 .ç + ÷ E cr σ c 0.2 è σ c 0 .2 ø

giving: σirr =

2008 0.002 . E cr . n r æ σ irr ö .ç 1+ ÷ σ c 0. 2 è σ c 0 .2 ø

( nr −1)

or even: σirr =

2008 12 .83 0.002 . 73800 . 13.83 æ σ irr ö .ç 1+ ÷ è 270 ø 420

after numerical resolution, the following is obtained for the panel


σirr = 412 MPa



V1-3 •

STIFFENERS

V1-3 •7-CRIPPLING Under a compression load, a stiffener may buckle locally (e.g.: local buckling of the flange). However, after local buckling, this stiffener may still support an increase in the load before failure occurs. The overload is supported by the stiffer areas not yet subjected to buckling. Crippling is the failure of the stiffener under a compression load.

The crippling load is the sum of the load-carrying capacity of the elements comprising the angle section. The cross section area of the stiffener is divided into single elements: 1

S2, σcri2 2

S3, σcri3

S1, σcri1

3

The critical crippling stress shall be calculated as follows: ΣS i σ cri σcrip = Eq. 1.3.7 ΣS i with: σcri: critical crippling stress of item i Si: area of the section of item i

Crippling - Page V1-3•7/1 revision 1


E S S E N T I A L


V1-3 •

STIFFENERS


Crippling - Page V1-3•7/2 revision 1



V1-3 •

STIFFENERS

V1-3 •4-WARPING V1-3 •4.1-ASSUMPTIONS Here the problem involved is the general buckling of the stiffener cap supported by an elastic foundation: the web. In fact, if the flange subjected to compression load buckles, it may buckle the entire stiffener. Therefore, the case will be considered as being between two stiffener cleats to isolate the flange as a beam, supported at its ends and elastically retained by the web over its entire length. The assumption taken is that the stiffener flange behaves like a beam with two hinges at the cleats. The cleats are considered as being rigid supports as defined in Chapter V-1-9 "Cleats".

et

bt Z CdGflange

ea ha

L

Y X

Fig. 1.3.4.1

Warping - Page V1-3•4/1 revision 1



V1-3 •

STIFFENERS

V1-3 •4.2-STIFFNESS OF THE EQUIVALENT ELASTIC MEDIUM

Fig. 1.3.4.2.1 Let "α" designate the elastic constant of equivalent discrete supports and "a" the equal distance separating two supports. The stiffness of the equivalent elastic medium is expressed by the quantity: α β= a Where "β" called the "foundation modulus" has the dimensions of a force divided by the square of a length. It represents the intensity of the foundation reaction. In this case, the bending rigidity around axis "x" of the web characterises the elastic foundation and is expressed as follows:

F =α * f ; 3. E c . I X. web α= h 3a a. e 3a IX.web = 12 (refer to Figs. 1.3.4.1 & 1.3.4.2.1)

Fig. 1.3.4.2.2




V1-3 •

STIFFENERS

The stiffness of the equivalent elastic medium is: E æe ö β = c .ç a ÷ 4 è ha ø

3

Eq. 1.3.4.2

with: Ec: ea: ha:

stiffener elasticity modulus web thickness web height (refer to Fig. 1.3.4.1)

V1-3 •4.3-WARPING STRESS (linear elasticity) The energy method is used to determine the critical buckling load (Ref. 1, paragraph 2.10) considering the deflection of the elastic line as a sinusoidal series with a wave node at each end of the line, i.e. at the stiffener cleats. By minimising this deformation energy, the lowest critical buckling load is obtained, which is: critical warping stress (linear elasticity): 2 ö β. L4 ~ = π . E c . I zt . æç m 2 + σ Eq. 1.3.4.3 ÷ dr 2 2 4 St L m . π . E c . I zt ø è with: Ec: stiffener elasticity modulus IZt: flange minimum inertia (*) St: area of the flange section (*) L: length of the beam modelling the flange between the cleats (refer to Fig. 1.3.4.1) m: number of wave half lengths β: elastic foundation modulus (refer to Eq. 1.3.4.2)

"m" (number of half wave lengths) shall be an integer and greater than or equal to 1 as the cleats are located at the wave nodes, in the assumptions. This number of wave half lengths "m" in which the bar is sub-divided during buckling is determined by the condition that the equation 1.3.4.3 is minimal. (*) IZt is the minimum inertia of the flange in relation to the Z axis, therefore passing through the centre of gravity of the flange section, which is expressed as follows: • case of a flange without a flanged edge: e t . b 3t Izt = (refer to Fig. 1.3.4.1) 12 • case of a flange with a flanged edge: Zt Izt = I(Gt;Zt) Gt (main inertia)




V1-3 •

STIFFENERS

~ V1-3 •4.4-DETERMINATION OF m0 MINIMISING σ dr ~ it is necessary to determine "m" such that: To minimise σ dr ~ dσ dr =0 dm

~ Determination of m giving the lowest values of σ dr β L m0 = 4 Eq. 1.3.4.4 π E c . I zt with: Ec: elasticity modulus of the stiffener in compression IZt: inertia of the flange in relation to the Z axis (refer to Fig. 1.3.4.1) L: length of the beam modelling the flange between cleats (refer to Fig. 1.3.4.1) m0: number of wave half lengths β: elastic foundation modulus (refer to Eq. 1.3.4.1) There is little chance of the value of m0, determined with Equation 1.3.4.4, being an integer and we have already established that it is necessary to have an integer value of m to have a ~ will be carried out for the two closest wave node at the cleats. Therefore, the calculation of σ dr ~ will integers bracketing the calculated m and the integer value of m giving the minimum σ dr be used. ~ calculated with m0 is the warping stress when the cleats are infinitely spaced apart σ dr ~ (m ) = E c σ dr 0 St

æe ö I zt ç a ÷ è ha ø

3

~ (E(m )) > σ ~ (E(m + 1)) then it is necessary to take m = E(m + 1) If σ dr 0 dr 0 0 otherwise m = E(m0) with E(m0) = integer part of m0




V1-3 •

STIFFENERS

~ V1-3 •4.5-CALCULATION OF σ dr Finally, putting down: ~ (m ) = 2 σ dr 0 St then:

E c I zt β , m 0 =

L π

4

β , m 1 = E( m 0 ) and m 2 = E( m 0 + 1) E c I zt

~ ~ (m ) = σ dr ( m 0 ) σ dr i 2

ææ m ö 2 æ m ö 2 ö çç i ÷ + ç 0 ÷ ÷ çè m ø è m i ø ÷ø 0 è

And the critical stress is calculated as follows: ~ = min σ~ ( m ), σ~ ( m ) Eq. 1.3.4.5 σ [ dr 1 dr 2 ] dr

V1-3 •4.6-PLASTICITY CORRECTION ~ , the critical stress calculated with η = 1, if: σ ~ ≥ 0.5 . σ then the critical Putting down σ dr dr c0.2 stress after plasticity correction shall be: ~ σdr = η5 σ Eq. 1.3.4.6 dr σ ~ This calculation is iterative as η5 = ƒ(ET, ν) = g(σdr). It converges when: dr → σ dr η5 η5 is the plasticity correction factor: E η5 = T with ET = tangent modulus Ec Ec = Young's modulus in compression of the stiffener In the design phase, we shall work in such a manner that σcr, (cor) ≥ σc0.2 is obtained, so that only σc0.2 is used as an allowable compression value.




V1-3 •

STIFFENERS





V1-3 •

STIFFENERS

V1-3•4.7-BLOCK DIAGRAM - SUMMARY

E

WARPING CALCULATION OF IZt

S

CALCULATION OF St Calculation of β E æe ö β = c .ç a ÷ 4 è ha ø

3

Eq. 1.3.4.2

S

Calculation of m0

m0 =

L π

4

β E c I zt

Eq. 1.3.4.3

E

~ (m ) Calculation of σ dr 0 2 ~ (m ) = E c I ztβ σ dr 0 St

Eq. 1.3.4.5

N

~ Calculation of σ dr with: m1 = E(m0) m2 = E(m0 + 1) ~ 2 2 æ m0 ö ö σ dr ( m 0 ) æ æ m i ö ~ ç ÷ σ dr (mi) = ç çè m ÷ø + çè m ÷ø ÷ 2 0 i è ø

Eq. 1.3.4.5

T

then:

~ = min σ~ ( m ), σ~ ( m ) σ [ dr 1 dr 2 ] dr

I

PLASTICITY CORRECTION no

~ ≥ σ dr

0.5 . σc0.2 ?

yes

~ σdr = η5 σ dr E Eq. 1.3.4.6 with: η = T 5 Ec

End Warping - Page V1-3•4/7 revision 1


A L


V1-3 •

STIFFENERS

V1-3 •4.8-EXAMPLE Using the stiffener studied in this chapter:

16 4

30

2 3

2 3

25 35

The characteristics are: (refer to paragraph V.1.3.1)

σr = 495 Mpa σc0.2 = 420 Mpa

Material: 7075 T7351 flange thickness: flange width: web thickness: web height:

et = 4 mm bt = 16 mm ea = 2 mm ha = 26,5 mm

Inter-frame distance:

L = 530 mm

Ec = 73800 MPa n = 13.83

V1-3 •4.8.1-Calculation of St and of IZt St = et . bt Izt =

St = 4 . 16

e t . b 3t 12

Izt =

St = 64 mm²

4 . 16 3 12

Izt = 1365,3 mm4

V1-3 •4.8.2-Foundation stiffness E æe ö β = ca . ç a ÷ 4 è ha ø

3

73800 æ 2 ö β= .ç ÷ è 26.5 ø 4

being therefore

3

giving β = 7.93 MPa

V1-3 •4.8.3-Calculation of m0 m0 =

L π

4

β E ct . I zt

giving


m0 =

530 π

4

7.93 73800 .1365.33 and therefore m0 = 2.83



V1-3 •

STIFFENERS

V1-3 •4.8.4-Critical stress (linear elasticity) ~ (m ) = 2 E I β •σ dr 0 c zt St • m1 = E(m0) • m2 = E(m0 + 1)

~ (m ) = 2 73800 . 1365.3 . 7.93 σ dr 0 64 m1 = 2 m2 = 3

~ (m ) = 883 MPa σ dr 0

2 2 ~ ~ (m ) = σ dr ( m 0 ) æç æç m i ö÷ + æç m 0 ö÷ ö÷ •σ dr i ç ÷

èè m 0 ø

2

è mi ø ø

2 2 ~ (m ) = 883 æç æç 2 ö÷ + æç 2.83 ö÷ ö÷ = 1104.5 MPa σ dr 1 è 2 ø ø 2 è è 2.83 ø 2 2 883 æ æ 3 ö 2.83 ö ö æ ~ σ dr (m2) = ÷ +ç ÷ ÷ = 889 MPa çç è 3 ø ø 2 è è 2.83 ø

~ (m ) < σ ~ (m ) then σ ~ =σ ~ (m = 3) As σ dr 2 dr 1 dr dr 2

~ = 889 MPa let: σ dr

V1-3 •4.8.5-Plasticity correction ~ = η .σ σ dr 5 dr with

η5 =

ET E ct

and

if σcr ≤

σ c0.2 then ET = Ec 2

otherwise ET =

In this case:

1 1 0.002 . n æ σ cr ,cor ö + ç ÷ Ec σ c 0.2 è σ c 0. 2 ø

n −1

~ = 889 MPa > σ c0.2 σ dr 2

To determine the corrected critical stresses, it is necessary to solve the following equations numerically. ~ σ dr σdr = n −1 0.002 . n E c æ σ dr ö 1+ ç ÷ σ c 0 .2 è σ c 0.2 ø 889 application: σdr = 12 .83 0.002 .13.83 . 73800 æ σ dr ö 1+ ÷ ç è 420 ø 420 after numerical resolution, the following is obtained σdr = 380 MPa




V1-3 •

STIFFENERS





YES

X

NO


1 4

5 4

2 1 2

Volume 1

5


Purpose

Scope


Contents

Document Manager



Not applicable.

V1 - 1 V1 - 2 V1 - 3 V1 - 4 V1 - 5 V1 - 6 V1 - 7 V1 - 8 V1 - 9



Validation

Name: J. HUET



3page 1

Title - Annex

Reference documents


Abbreviations




Definitions


Highlights

Issue

Date

Pages modified

A

02/98

V1 - 1 à V1 - 3 V1 - 7 à V1 - 9

B

05/99

V1 - 7


V1 - 4

New chapter.

V1 - 1


C

11/99




New chapter.

MTS 004 Iss. C

3Ann. page



List of approval

Dept. code

Function

BTE/CC/CM

Name / First name


Key words

Calcul

Bibliography

Néant

Signature

CAZET G.

Distribution list

Dept. code

Function


BQP/TE

Archives Diderot

SIBADE Alain

BQP/TE


SIBADE Alain


BOUTET Fernand

BTE/SM/MG



MTS 004 Iss. C

page IG1

MCS V1-4 • THIN WEB BEAMS CONTENTS

V1-4 V1-4 V1-4 V1-4 V1-4 V1-4 V1-4 V1-4 V1-4 V1-4 V1-4

THIN WEB BEAMS 1 GENERAL 2 PREPARATION OF BASIC DATA 3 GENERAL PRINCIPLES FOR DIAGONAL TENSION 4 WEB FOLDING STRESS 5 STUDY OF UPRIGHTS IN STANDARD ZONE 6 STUDY OF FLANGES 7 END SPANS AND SPANS WITH OPENINGS 8 STUDY OF WEB 9 STUDY OF FASTENERS 10 EXAMPLE

V1-4 1

issue

date

revision

0 0 0 0

01/1999 01/1999 01/1999 01/1999

Original issue Original issue Original issue Original issue

0 0 0 0 0 0 0

01/1999 01/1999 01/1999 01/1999 01/1999 01/1999 01/1999

Original issue Original issue Original issue Original issue Original issue Original issue Original issue

GENERAL

1–1

V1-4 1.1 DEFINITIONS V1-4 1.1.1 Thin web beams V1-4 1.1.2 Pocket and super-stiffener

1–1 1–1 1–2

V1-4 1.2 BEHAVIOUR OF A THIN WEB BEAM

1–3

V1-4 1.3 FAILURE MODES

1–4

V1-4 1.4 SCOPE, LIMITS

1–6

V1-4 2

PREPARATION OF BASIC DATA

2–1

V1-4 2.1 MATERIAL CHARACTERISTICS

2–1

V1-4 2.2 SECTION OF A SUPER-UPRIGHT IN STANDARD ZONE

2–1

V1-4 2.3 END SUPER-UPRIGHT SECTIONS

2–3

V1-4 2.4 POCKET DIMENSIONS, BOUNDARY CONDITIONS

2–4

V1-4 3

GENERAL PRINCIPLES FOR DIAGONAL TENSION

3–1

V1-4 3.1 INCOMPLETE DIAGONAL TENSION THEORY

3–1

V1-4 3.2 LOADING RATIO, DIAGONAL TENSION FACTOR OF A POCKET

3–1

V1-4 3.3 LOAD-CARRYING WIDTH

3–2

V1-4 4

WEB FOLDING STRESS

Issue 0 © AEROSPATIALE - 1999

Contents

4–1

page V1-4•i

MCS V1-4 • THIN WEB BEAMS V1-4 5

STUDY OF UPRIGHTS IN STANDARD ZONE

5–1

V1-4 5.1 GENERAL

5–1

V1-4 5.2 LOAD-CARRYING WIDTHS AND SECTIONS

5–2

V1-4 V1-4 V1-4 V1-4

5.3 STRESSES IN UPRIGHTS 5.3.1 General method 5.3.2 Fold angle 5.3.3 Simplified approaches

5–3 5–3 5–4 5–4

V1-4 V1-4 V1-4 V1-4

5.4 ALLOWABLE STRESSES IN UPRIGHTS 5.4.1 column buckling (refer to V1-1.5.9.3) 5.4.2 forced crippling 5.4.3 natural local buckling

5–5 5–5 5–5 5–5

V1-4 6 V1-4 V1-4 V1-4 V1-4 V1-4 V1-4 V1-4 V1-4 V1-4

STUDY OF FLANGES

6–1

6.1 LOADS AND STRESSES IN FLANGES 6.1.1 beam geometrical characteristics 6.1.2 loads and stresses due to primary bending of beam 6.1.3 additional loads and stresses due to the action of the tension field, mean strain 6.1.4 total normal loads and mean stresses in flanges 6.1.5 secondary bending moment in flanges 6.1.6 stresses due to secondary bending moment in flanges 6.1.7 overall stresses in flanges 6.1.8 allowable stresses and margins

6–1 6–1 6–2 6–2 6–3 6–3 6–4 6–4 6–5

V1-4 6.2 EXAMPLE

6–6

V1-4 6.3 CONCLUSION

6–10

V1-4 7

END SPANS AND SPANS WITH OPENINGS

7–1

V1-4 7.1 GENERAL

7–1

V1-4 7.2 EDGE ELEMENTS

7–1

V1-4 7.3 EXAMPLE

7–2

V1-4 7.4 ADDITIONAL STUDIES FOR OPENINGS IN A THIN WEB BEAM

7–4

V1-4 8

STUDY OF WEB

8–1

V1-4 8.1 STRESSES IN WEB V1-4 8.1.1 In pocket thicknesses V1-4 8.1.2 At uprights

8–1 8–1 8–1

V1-4 8.2 WEB FAILURE

8–1

page V1-4•ii © AEROSPATIALE - 1999

Contents

Issue 0

MCS V1-4 • THIN WEB BEAMS V1-4 9

STUDY OF FASTENERS

9–1

V1-4 9.1 WEB-FLANGE AND WEB-END UPRIGHT CONNECTIONS

9–1

V1-4 9.2 STANDARD DISSYMMETRICAL WEB-UPRIGHT CONNECTIONS (SINGLE ANGLE)

9–2

V1-4 9.3 UPRIGHT-WEB CONNECTIONS

9–2

V1-4 10

EXAMPLE


10–1

Contents

page V1-4•iii

MCS V1-4 • THIN WEB BEAMS SYMBOLS h:

Beam height.

h1:

Beam height between fastener rows.

h2:

Beam height between flange CGs.

hm:

Upright height between centre of rivets attaching upright to flanges.

I:

Inertia (index 0: initial inertia before folding).

B,Bi: Distances between uprights.

Is:

Flange inertia.

UL: Ultimate loads.

Iss:

Upper flange inertia.

LL: Limit loads.

Isi:

Lower flange inertia.

CS: Single shear.

Iaa:

Beam inertia with web.

Cfs: Attenuation factor.

Isa:

Beam inertia without web.

d:

CG (centre of gravity) offset.

IXXa: Inertia of web around an XX axis.

d a:

Offset between web CG and total section CG.

IXXm: Inertia of upright around an XX axis.

dm:

Offset of upright CG and total section CG.

k,ki:

Diagonal tension factors.

D:

Fastener diameter.

Ks:

Shear buckling coefficient.

l:

Distance between load introduction point and section studied. Reduced effective length.

a:

Distance between end rows of web-flange fasteners.

b,bi: Distances between end rows of web-upright fasteners. b c:

Width of free flange of an angle.

A:

Distance between flanges.

e, ei: Web thickness on either side of an upright. es:

Member flange thickness.

Le:

esi:

Lower flange thickness.

Lt,Lti: Load-carrying widths.

ess:

Upper flange thickness.

Lt0:

et: e’:

Initial load-carrying width (before folding).

TD

Land thickness.

Lt : Load-carrying width after folding.

Equivalent thickness.

M f:

E:

Young's modulus (indexes "a": web, "m": upright). Mf1:

Es:

Secant modulus.

Et:

Tangent modulus.

fb: fr:

Bending moment. Bending moment concerning beam resisting shear.

Mf2:

Bending moment concerning beam resisting pure diagonal tension.

Linear load inducing bending in edge elements.

Mfs:

Secondary bending moment.

Load per length unit acting on the rivets.

My:

Bending moment around y-axis.

n:

Work hardening coefficient (Ramberg and Osgood).

fmax: Max shear flow. F1:

Mean axial load in a flange due to shear.

F2:

Mean axial load in a flange due to pure diagonal tension.

N:

Normal load (index x or y => along x or y).

p:

Rivet pitch.

Compression load induced by shear load under action of the tension field.

Pm:

Load transferred from base to upright.

Fc:

q:

Length of an opening.

Fsim: Mean axial load in lower flange.

Ri:

Reactions at flanges.

Fssm: Mean axial load in upper flange.

Rs:

Loading ratio in single shear.

Fsit: Mean total axial load in lower flange.

S:

Section (index 0: initial section before folding).

Fsst: Mean total axial load in upper flange.

Sa:

Web section.

g:

Distance between web centre plane and upright CG.

Ss:

Flange section.

Sss:

Upper flange section.

G:

Shear modulus (Coulomb).

Ssi:

Lower flange section.

Gs:

Secant shear modulus.

page V1-4•iv © AEROSPATIALE - 1999

Contents

Issue 0

MCS V1-4 • THIN WEB BEAMS Saa:

Section of beam with web.

Ssa:

Section of beam without web.

Sm:

Section of upright alone.

Sequi: Equivalent section. T:

Shear load.

Tcis:

Shear load acting in form of pure shear.

TTD:

Shear load acting in form of pure diagonal tension.

TD:

Diagonal tension.

σ0.2:

Conventional yield strength (indices "a": web, "m": upright).

σR:

Breaking tensile stress (indices "a": web, "m": upright).

σ TD x : Mean stress along a flange. σTD x min :Minimum stress along a flange. σ TD y : Mean stress along an upright. σTD y min :Minimum stress along an upright. σa:

Normal stress in web.

TDI: Incomplete diagonal tension.

σxa:

Normal stress along x-axis in web.

vi:

σya:

Normal stress along y-axis in web.

σflf:

Allowable forced crippling stress.

Dimensions between neutral fibre of a flange and its extreme fibres. Static moment (index 0: static moment before folding).

σ fss : Primary bending stress at extreme fibre of upper flange.

Parameter= 0.7 * b * ê ú ë (I ss + I si ) * h 2 û W sem: Static moment around neutral fibre of flange.

σ fsi : Primary bending stress at extreme fibre of lower flange.

W: W d:

σssm: Mean stress in upper flange due to primary bending.

W ame: Static moment of web portion above neutral fibre. W XXa: Static moment of web around XX axis. W XXm: Static moment of upright around XX axis. yi:

Dimensions between neutral fibre of beam (with or without web) and extreme fibres of flanges (upper or lower).

σsim:

Mean stress in lower flange due to primary bending.

σssc:

Compression stress in upper flange due to tension field.

σsic:

Compression stress in lower flange due to tension field.

σsst:

Total upper flange mean stresses.

σsit:

Total lower flange mean stresses.

σfsi:

Secondary bending stresses in extreme fibres (upper or lower) of flanges (upper or lower).

α:

Fold angle in pockets.

ε:

Strain (index x or y => along x or y).

εTD:

Diagonal tension strain.

Φy:

Transverse flow.

η:

Plasticity correction coefficient.

σssTi: Overall stress in lower fibre of upper flange.

λ:

Slenderness ratio.

σssTs: Overall stress in upper fibre of upper flange.

νe:

Poisson ratio – elastic.

σsiTi: Overall stress in lower fibre of lower flange.

νp:

Poisson ratio – plastic.

σsiTs: Overall stress in upper fibre of lower flange.

ν:

Poisson ratio – elasto-plastic.

τ,τi:

Shear stresses.

ρ:

Radius of gyration of a section.

τcr:

Critical stress.

σ:

Normal stress (index x or y => along x or y).

τcr,a:

Critical stress in web.

σ0:

Allowable compression stress with zero slenderness ratio.


Contents

page V1-4•v

MCS V1-4 • THIN WEB BEAMS BIBLIOGRAPHY

1-

KUHN & PETERSON & LEVIN METHOD OF ANALYSIS : PLANE - WEB SYSTEMS. NACA T.N 2661 , MAI 1952.

2-

BRUHN ANALYSIS & DESIGN OF FLIGHT VEHICLE STRUCTURES

page V1-4•vi © AEROSPATIALE - 1999

General

Issue 0

MCS V1-4 • THIN WEB BEAMS V1-4 1 GENERAL The aeronautical structure designer is constantly searching for types of structure or calculation methods enabling a gain in weight whilst meeting manufacturing and cost constraints. A good example is that of thin beams where the buckling of the web is accepted, the shear loads being balanced by the web diagonal tensile stresses. The rules for calculating these beams are derived from the principles established for the stiffened panels in chapter V1-1.

V1-4 1.1 DEFINITIONS V1-4 1.1.1 Thin web beams We call "thin web beams" beams where the webs are subjected to diagonal tension (*). In general, they are distinguished by: - concentrated longitudinal members called flanges which, practically alone, transmit all normal (tensile or compression) and bending loads, - a thin web the only practical role of which is to ensure the transmission of the shear load by diagonal tension and shear which induces a compression overload in the flanges and uprights, - transverse members called uprights the function of which is to provide supports, more or less regularly spaced and more or less rigid, for the flanges and the web. The uprights balance the diagonal tension. This definition establishing the respective roles of the web and the flanges concerning load transmission in reality comprises only an "ideal limit". Indeed, we know, from chapter V1-1, that all thin webs participate in the transmission of the normal loads until its stability limit is reached. We also know that beyond this limit a thin web still participates but to a lower extent evidenced by the "load-carrying" width notion. Also, after folding, and whilst continuing to transmit the shear load, the thin web is subjected to strains higher than those defined by the linear elasticity theory. Indeed, the efficient transverse modulus of elasticity GTDI, representing the rigidity of the skin to shear in the presence of diagonal tension, decreases as the load increases (refer to § V1-1.5.10). Thin web beams are significantly more flexible than stable web beams. This marks the limit of their applications. We learnt, in chapter V1-1, how to evaluate the "additional loads" consecutive to the instability of stiffened panels. These theories apply generally to all the beams studied below for which we admit the buckling of the web. The interest of these thin web beams lies in their lightness and flexibility. The stiffer beams where the buckling of the web is not tolerated are covered by chapter V1-5. Flange

Web Uprights Flange

(*): The diagonal tension theory is described in chapter V1-1.5


General

page V1-4•1–1

MCS V1-4 • THIN WEB BEAMS V1-4 1.1.2 Pocket and super-stiffener (Refer to §V1-1.1.1 ) The definition of the thin web beams implies that the beam is criss-crossed by longitudinal and transverse members. These are attached to the web either because they are "sewn" by means of screws, rivets or tack welding ("added flanges or uprights"), or because the assembly is machined from a plate ("integrated flanges or uprights"). We call: - "pockets": the web quadrilaterals delimited by the members. - "super-stiffener": the association of an upright or a flange with an associated web portion (half-pockets on either side).

Members

Pocket thickness

Land

B

B

Uprights

page V1-4•1–2 © AEROSPATIALE - 1999

General

Issue 0

MCS V1-4 • THIN WEB BEAMS V1-4 1.2 BEHAVIOUR OF A THIN WEB BEAM x T

y Mf a

a

The external loads which can be applied to a thin web beam are: • T:

Shear load

• Mf: Bending moment Induced effects: These loads mainly generate in the beam normal internal loads in the flanges and the uprights and a shear and diagonal tension flow in the web of the beam. Below, we show a section of the thin web beam at an upright, representing it associated with a load-carrying web portion. The internal loads applied to this section are: Nx: Normal load in the upright and its load-carrying web portion, induced by diagonal tension.

Φy: The normal transverse flow due to diagonal tension in the web. The stresses resulting from these loads are: Normal (σx) in the upright. Normal (σx), transverse (σy) and tangential (τxy) in the web.

Section a-a

Upright

Nx

z

Φy G

y

Web

Φy

x Remark: In certain specific cases, secondary induced loads are added to the loads applied above (e.g.: transfer of neutral fibre to an upright runout induces a bending moment My in the upright).


General

page V1-4•1–3

MCS V1-4 • THIN WEB BEAMS The calculating difficulties come from the fact that the behaviour of the beam depends on the loading intensity.

• Tension, the static strength limit at ultimate load is given by the characteristics of the material. • Shear, the same applies even though the folds occur beyond the critical shear of the pockets. These folds are oblique to the edges. For this reason, the term "diagonal tension" is used even though inappropriate. • This property of dependence is further emphasised in the case of compression. It is related to the instability of the thin web.

V1-4 1.3 FAILURE MODES There are several types of instability and failure.

• Web instability: Thin sheets fold under relatively low compression and/or shear loads. On the compressed flange side, the web is saturated when its buckling stress is reached. At greater values, the excess load is picked up by the flange and the load-carrying portion of the web. In shear cases, after the appearance of the folds, the web transfers the excess load partly in the form of shear stresses and the remainder as diagonal tension. The tangential flows at pocket edges are the same at constant loads with or without the folds. However, the diagonal tension induces flows normal to these edges which causes overloads, especially in the flanges and uprights. The instability of the web does not lead to the failure of the beam and is therefore not a cause for limiting the ultimate load.

• Instability of super-stiffeners If the longitudinal or transverse members, spaced respectively at a pitch of A or B, are sufficiently rigid, the superstiffeners (upright or flange + load-carrying web) behave as columns of length A or B on single supports. Unstable failure may occur in the most critical super-stiffener under: - local buckling, - column buckling, - forced crippling of the flanges of the members (inter-rivet buckling does not exist when the web is subject to diagonal tension).

• Failure of the skin Generally, the web is thicker at the members (uprights, flanges) than in the pockets. Failure can then occur, either in the pocket-land blend-in radii or in the web-members fastener rows. It must be checked that the maximum shear in the web does not exceed half the material tensile breaking stress (TRESCA criterion).

• Failure of fasteners Under the combined effect of shear and diagonal tension in the web. Also, the secondary effects must be taken into account: - Bending of the flanges due to diagonal tension, - Secondary bending of the dissymmetrical uprights.


General

Issue 0

MCS V1-4 • THIN WEB BEAMS Bent flange (1)

Row of fasteners (4) Section a-a

Section b-b

e

A

Opening in web (2)

Upright run-out (3)

Section c-c

M1

(1) Secondary bending of the flanges due to diagonal tension.

M1

(3) Secondary bending of the dissymmetrical uprights Section b-b

(4) Fasteners flange web

M2

F1 F2

R

F1 due to the shear flux in the web. F2 due to the transverse stress in the web generated by resulting diagonal tension R. Secondary bending of the uprights

N

N TD upright

d

d

Web stresses -N - Nd

N TD upright Equilibrium: TD N TD x web + N x upright = 0 d N TD x upright + M f upright = 0


General

page V1-4•1–5

MCS V1-4 • THIN WEB BEAMS (1): Bent flange Each flange is forced towards the inside of the web by a load due to the diagonal tension. The web behaves as a beam clamped at each end at the uprights. (2): Calculating holes and openings The presence in a thin web beam of flanged holes or openings of any sort creates overstresses in the sheet around the holes whereas secondary bending loads appear in the flanges. (3): Upright run-out The run-out of an upright overloads the upright by a secondary moment due to the transfer of the neutral fibre between the web and the centre of gravity of the upright. (4): Fasteners After folding, the web pulls on its edges, hence additional loads perpendicular to the alignment of the fasteners are to be taken into account. (5): Web The bending of the flanges leads to a variation in the tension field in the pockets. Their loads are increased in the vicinity of the flange and upright intersections.

V1-4 1.4 SCOPE, LIMITS SCOPE In addition to the thin web beam calculation, this chapter deals with certain specific cases relevant to stiffened panels which were not studied in chapter V1-1: - bending of flanges due to diagonal tension or bending of a stiffener bordering on an opening in a stiffened panel, - influence of holes in the web or in the skin of a stiffened panel, - effect induced by a stiffener run-out (secondary bending), - calculation of the fasteners attaching the web to the flanges or attaching the stiffened panels together, - sizing of the fasteners attaching the web to the uprights or attaching the stiffeners and the skin of a stiffened panel. LIMITS

es ≥ 0.6 (the flange of the members on the skin side must not be too thin) e B A ≤ 1; 120≤ ≤ 1500 (the proportions of the pockets must be within these ranges) 0.2 ≤ A e p 2.5 ≤ ≤ 10 (riveting) D Caution: risk of non-conservative predictions with thick pockets (outside range). The uprights are more or less equally spaced and thicknesses of adjacent pockets are similar.


General

Issue 0

MCS V1-4 • THIN WEB BEAMS V1-4 2 PREPARATION OF BASIC DATA V1-4 2.1 MATERIAL CHARACTERISTICS STABILITY CALCULATIONS:

The elasto-plastic behaviour of the materials must be taken into account in these calculations. To this end, the Ramberg and Osgood model (refer to V2-2 "Material behaviour") is used:

ε=

æ σ ö σ ÷÷ + 0.002 çç E è σ0.2 ø

Es =

n

σ ε

1 n 1− n = + E t Es E

Therefore, for each material used, the following is required: • Young's modulus of compression, E, • Compression yield strength, σ0.2 • R.&.O. coefficient, n.

V1-4 2.2 SECTION OF A SUPER-UPRIGHT IN STANDARD ZONE SECTION (INTEGRATED UPRIGHT)

Lt2

Lt1

e2 d

G e1

Z

Initial section: Lti = Bi/2

CHARACTERISTICS: Cross section area: Moment of intertia/GX: C.G. offset: Load-carrying width: Pocket thicknesses:


X

et

B1

B2

S I d Lt1, Lt2 e1, e2

Preparation Of Basic Data

page V1-4•2–1

MCS V1-4 • THIN WEB BEAMS SECTION (ADDED UPRIGHT):

Lt2

Lt1

e2 X

X

da d

e1

G

dm

Z

Initial section: Lti = bi/2

b1

b2

b1 and b2 are the distances between outer fastener rows. WEB CHARACTERISTICS

MEMBER CHARACTERISTICS (beam upright)

Cross section area: Moment of inertia/XX: Static moment/XX: Load-carrying width: Pocket thicknesses:

Sm IXXm W XXm

Sa IXXa W XXa Lt1, Lt2 e1, e2 W da = XXa Sa

C.G. offset:

da =

W XXm Sm

Remark: W XXa is negative whereas W XXm is positive. CHARACTERISTICS OF A SUPER-STIFFENER MADE SYMMETRICAL: If a super-stiffener is not symmetrical in relation to the (G, Y, Z) plane, an equivalent symmetrical super-stiffener must be constructed.

Lt2.

Lt1.

e2 X

da e1

d

G

X

dm Index 0 is relevant to the initial section. Z

e e e1Lt1.0 + e 2 Lt 2.0 e τ + e 2τ2 e = 1 2 ; Lt :τ= 1 1 0= 2 e1 + e 2 e1 + e 2 Diagonal tension factor: k =


æ1 + k ö k1 + k 2 ; Loading ratio: R s =ç ç 1 − k ÷÷ 2 è ø

2.3


Issue 0

MCS V1-4 • THIN WEB BEAMS V1-4 2.3 END SUPER-UPRIGHT SECTIONS SECTION (INTEGRATED UPRIGHT)

B Lt

d

e

X G

End stiffener

Z

Initial section: Lt = B/2

CHARACTERISTICS: Cross section area: Moment of inertia/GX: C.G. offset: Load-carrying width: Pocket thickness:

S I d Lt e

SECTION (ADDED UPRIGHT): Initial section: Lt = b/2

B Lt et

da

X e

d dm

X G

Z b: is the distance between outer fastener rows. WEB CHARACTERISTICS Cross section area: Moment of inertia/XX: Static moment/XX: Load-carrying width: Pocket thickness: C.G. offset:

MEMBER CHARACTERISTICS (beam upright) Sm IXXm W XXm

Sa IXXa W XXa Lt e W da = XXa Sa

dm =

W XXm Sm

Remark: W XXa is negative whereas W XXm is positive.



page V1-4•2–3

MCS V1-4 • THIN WEB BEAMS V1-4 2.4 POCKET DIMENSIONS, BOUNDARY CONDITIONS Use the conditions given in § V1-1.3 (pocket folding stresses in stiffened panels): If the members have opened profiles, the edge conditions to be chosen for the pocket thickness buckling stress calculation are: four hinged edges. b' (distance between fasteners) et e b"

B

Width of pocket "b" is: b = B for integrated stiffeners. b = b’ for added stiffeners. b = b’’ for all types, ift ≥e 3e Height of pocket "a" is: a = A, a’, a’’ (same principle as above)



Issue 0

MCS V1-4 • THIN WEB BEAMS V1-4 3 GENERAL PRINCIPLES FOR DIAGONAL TENSION V1-4 3.1 INCOMPLETE DIAGONAL TENSION THEORY This theory is detailed in §V 1-1.5.1

Let us recall the main principle: In a normal thin web beam, the stress condition in the web is between the pure shear stress condition and the pure diagonal tension condition. This theory is based on the assumption that the total shear load in the web can be broken down into two parts: T = Tcis + TTD - a pure shear part Tcis = (1 - k)T (where k is the diagonal tension factor) - a pure diagonal tension part TTD = kT. y

x

Web not buckled: pure shear

τ

Web subjected to pure diagonal tension

σy = kτ tan α

- (1 - k)τ TD

TDI α

+ 2 kτ

τ

CS (1 - k)τ

45°

σx =

sin 2α

TDI

σ2 = - (1 - k) τ sin2α TDI

or

kτ

α

tan α

In base (x,y)

σ1 =

τ12 = (1 - k) τ cos2α 2 kτ sin α

+ (1 − k ) τ sin 2α

Facette along theoretical plies

V1-4 3.2 LOADING RATIO, DIAGONAL TENSION FACTOR OF A POCKET The loading ratio Rs and the diagonal tension factor k are defined by: Rs =

τ τcr , a

(τcr,a: critical shear in web)

k = 0 if Rs < 1 k=

R s0.4343 − 1 if Rs > 1 R s0.4343 + 1

Dissymmetrical pockets: refer to V1-4.2.2


General Principles For Diagonal Tension

page V1-4•3–1

MCS V1-4 • THIN WEB BEAMS V1-4 3.3 LOAD-CARRYING WIDTH UPRIGHT ORIENTATION σya σp

LtTD

LtTD

e Lt0

Lt

TD

= (1 − k)

Lt 0 =

Lt0 2

Lt0

Index 0 is relevant to the initial section (before folding)

b 2

FLANGE ORIENTATION σx σp

a e

Lt0

LtTD

Lt

TD

Lt 0 =

= (1 − k)

Lt0 2

Index 0 is relevant to the initial section (before folding)

a 2


General Principles For Diagonal Tension

Issue 0

MCS V1-4 • THIN WEB BEAMS V1-4 4 WEB FOLDING STRESS The folding stress of the web between two uprights is calculated according to the rules given in chapter V1-2 (Buckling of plates and thin shells): τcr,a = η ks: b: a: e: η:

æ ö e π2 ÷÷ k s E çç 2 12 (1 − ν e ) è min {a , b} ø

2

(Refer to § V1.2.2)

theoretical buckling coefficient for a panel the edges of which simply supported. width of sheet between upright-web fastener rows. height of web between flange-web fastener rows. web thickness. plasticity correction coefficient (Refer to § V1-2.1.3 and § V1-2.2.2)

Hinged edges: 2

æ ì a b üö ks = 3.8 çç min í ; ý ÷÷ + 5.35 î b a þø è G æ 1 + νe ö Es η= s = ç ÷ G è1+ ν ø E b

a

e

T

The dimensions of the pockets a, b and the boundary conditions to be taken into account are defined in § V1-4.2.4.


Web Folding Stress

page V1-4•4–1

MCS V1-4 • THIN WEB BEAMS

Page intentionally left blank


Web Folding Stress

Issue 0

MCS V1-4 • THIN WEB BEAMS V1-4 5 STUDY OF UPRIGHTS IN STANDARD ZONE V1-4 5.1 GENERAL

T

A

a

a

e

b

b

B

Symmetrical upright Web

Section a-a

Dissymmetrical upright Web

Section b-b

Two types of uprights can be identified on thin web beams: • symmetrical uprights formed of two identical sections on either side of the web (section a-a), • single upright on one side of the web (section b-b). Calculation of standard uprights is almost identical to that of the stiffeners of flat stiffened panels subjected to shear (Refer to § V1-1.5 ).


Study Of Uprights In Standard Zone

page V1-4•5–1

MCS V1-4 • THIN WEB BEAMS V1-4 5.2 LOAD-CARRYING WIDTHS AND SECTIONS σya

LtTD

σa

Upright

LtTD

Upright

Upright Flange

α

e X

X Lt0

Lt0

y x

Flange

Z

After the section of an upright has been made symmetrical in relation to the Z-axis (if necessary) as described in §.V1-4.2.2, the widths, sections and other load-carrying geometrical characteristics are given by the following formulas: Index 0 is relevant to the initial section, that is before folding. Lt • Load-carrying width LtTD = (1 - k) 0 2 • Load-carrying section let S0 be the initial total section and ∆STD the non load-carrying portion of the web (not shaded) ∆STD = 2 (Lt0 - LtTD) e = (1 + k) Lt0e STD = S0 - ∆STD • INTEGRATED UPRIGHT eö æ ∆W TD = ç d 0 − ÷ ∆STD 2ø è TD W = W0 - ∆W TD ∆W TD dTD = d0 + TD S 2 æ eö e2 ö æ ∆ITD = ç ç d 0 − ÷ + ÷ ∆STD çè 2ø 12 ÷ø è TD TD I = I0 - ∆I - (d0 - dTD)2 STD • ADDED UPRIGHT

CHARACTERISTICS OF BEAM WEB eö æ TD W XXa = W XXa0 + ç e t − ÷ ∆STD 2 è ø 2 æ eö e 2 ö TD çæ I TD ÷ + ÷ ∆S XXa = I XXa0 − ç e t − çè 2ø 12 ÷ø è CHARACTERISTICS OF CORRECTED SECTION (DIFFERENT MATERIALS) STD Sm Em E = aTD E a + TD S S E E TD TD W XX = a W XXa + m W XXm E E TD W XX dTD = TD S E a TD E I TD I XXa + m I XXm XX = E E TD ITD = I TD d TD XX − S


2


Issue 0

MCS V1-4 • THIN WEB BEAMS V1-4 5.3 STRESSES IN UPRIGHTS V1-4 5.3.1 General method The stress in an upright is not uniform: - Single uprights, dissymmetrical in relation to the web mean plane are the seat of secondary bending (Refer to § 1.3). - The stresses vary along the upright from a minimum at the beam flanges to a maximum in the centre of the upright ("gusset effect") (fig “b“). σ TD y σy

a

a

σ TD y min σy

Neutral axis Gusset effect σ TD y

σ TD y

Section a-a figure "a"

figure "b"

SECONDARY BENDING Due to the secondary bending caused by loading outside of the neutral fibre, compression stress is maximum at beam web (fig "a"). The "mean" maximum compression stress (variation σofalong y is not taken into account here) is determined as if it were a single compression by dividing the compression load by a "equivalent section", STD equi : σ TD = y

N TD y STD equi

where:

N TD = - 2 k Lt0 e τ tanα y

and

STD equi =

STD

ægö 1 + çç ÷÷ èρø

2

e 2 e Added upright: g = dTD + et 2 Integrated upright: g = dTD -

ρ=

I TD STD

α is the fold angle (See next page) TD Remark: If the upright is symmetrical, then g = 0 and STD equi = S GUSSET EFFECT The maximum compression is given by the following equation: TD σ TD σ TD b ö σ y min æ y min y min <1 = ( 1 − k ) 1 . 78 − 0 . 64 ; ≥ 1 ; take 1 if ç ÷ TD TD σ TD a ø σy σy è y a: height of the upright between the centres of the rivets attaching the upright to the flanges.



page V1-4•5–3

MCS V1-4 • THIN WEB BEAMS V1-4 5.3.2 Fold angle The rules for determining the α's described in § V1-1.5.6 (flat stiffened panels - stresses in the super-stiffener) are applied as follows: ε − ε TD x tan2α = TD ε TD − ε TD y

é 2k ù τ The elongation of the web along the folds is equal to: ε TD = ê + (1 − k ) (1 + ν ) sin 2α ú ë sin 2α û Ea TD The mean compression strains induced by the diagonal tension in the upright ( ε TD y ) and the flanges ( ε x ):

ε TD = y

σ TD y Ey

is described in § 6 (Study of flanges). Calculation of ε TD x

V1-4 5.3.3 Simplified approaches As the fold angle is around 45°, we suppose that sin 2α is equal to 1. τ Hence: εTD = [(1 + k ) + ν(1 − k )] Ea ASSUMPTION 1: If the stiffening ratio and the materials of the uprights are similar to those of the longitudinal members (the flanges), we consider that the folds are 45° irrespective of k. ASSUMPTION 2: If the longitudinal members (the flanges) can be considered as infinitely rigid, determine σ TD y and τ using the graph below.

Remark: This assumption is not conservative for the upright as the folds tend to bend over onto the more rigid members.

STD equi

STD equi

− 1 on the y-axis. If necessary, interpolate to obtain sought value. eb is the initial geometrical section (∆STD = 0).

Place k on the x-axis and

(

)

0.10 0.15

0.20

(

)

TD Analysis of diagonal tension: σ TD y / τ versus k and Sequi − be / be

1.0 σ/τ = 0.9

0.05

0.25

0.30 0.35

0.40

0.45 0.50 0.55 0.60

0.70

0.80

0.90 1.00

0.8

1.10

0.7

1.20

0.6

1.40

0.5

1.60 1.80 2.00

0.4 0.3

2.50 3.00 4.00 5.00

0.2 0.1 0.0 0

0.1


0.2

0.3

0.4

0.5 k

0.6


0.7

0.8

0.9

1

Issue 0

MCS V1-4 • THIN WEB BEAMS V1-4 5.4 ALLOWABLE STRESSES IN UPRIGHTS 3 types of failures can be encountered: 1. Column buckling. 2. Forced crippling. 3. Natural local buckling.

V1-4 5.4.1 column buckling (refer to V1-1.5.9.3) • Resistance to column buckling is checked, in principle, by comparing the mean stress at the centre of gravity of the super-upright ( σ yTD ) and the critical buckling stress. • When upright bending is initiated, the diagonal tension tends to oppose it. This interface action is taken into account by use of the clamping coefficient below: TD = LTD = KTDA; K

1 Lt ö æ 1 + k ç3 − 4 0 ÷ A ø è 2

TD TD TD Then calculate σ TD y crit from (S ; I ; L ) using the rules given in § V1-1.4 (the simplest method consists in using the Johnson formula).

• According to the slenderness ratio of the super-upright, the column may buckle under a two half-wave mode (especially dissymmetrical uprights). In addition, it must be checked that the maximum compression stress at the centre of gravity of the super-upright is LTD lower than the buckling stress calculated with a buckling length equal to : 2 é σ y min ù TD ú σ y ≤ σ æ LTD ö ê ÷ y crit ç ëê σ y ûú ç 2 ÷ ø

è

V1-4 5.4.2 forced crippling Ripples in the web favour the buckling of the flange of the upright attached to the web especially if the flange of the upright is thinner than the web. The allowable stress is calculated by means of the rules established for the stiffeners (refer to V1-3.5): σ flf = − 0.051

σ c 0.2 m

æ e' E k çç sm cm è e a E ca + 0.002 2 3

1

ö3 ÷ (indexes => c: compression; m: upright; a: web; s: flange) ÷ ø

σ c 0.2 m E cm e' sm is the equivalent thickness of the flange. TD On account of the local nature of this failure, stress σ TD y min (and not σ y ) must not exceed σflf.

V1-4 5.4.3 natural local buckling To avoid natural local buckling, the stresses in the upright must be lower than the local buckling stresses calculated by means of the rules in chapter V1-3. As added dissymmetrical uprights are the seat of secondary bending, they are practically limited only by the Forced crippling. The stresses to be compared with the allowable local buckling are those at the centre of the uprights (on account of the gusset effect).



page V1-4•5–5





Issue 0

MCS V1-4 • THIN WEB BEAMS V1-4 6 STUDY OF FLANGES V1-4 6.1 LOADS AND STRESSES IN FLANGES The loads in the flanges are: 1) The normal loads due to the primary bending of the beam calculated by the elementary bending theory. 2) The additional loads induced by the tension field in the web. 3) The secondary bending loads generated by the tension field (figure "a"). T

figure "a"

V1-4 6.1.1 beam geometrical characteristics l

T

A

e

b

h1 h2

h

A v1 v2

ess

v3

esi

Y3

y1

neutral axis

v4 neutral axis

Sss = upper flange section. Ssi = lower flange section. Iss = upper flange inertia. Isi = lower flange inertia. v1 = distance between upper upper flange and neutral axis. v2 = distance between lower upper flange and neutral axis. v3 = distance between upper lower flange and neutral axis. v4 = distance between lower lower flange and neutral axis. ess = upper flange thickness. esi = lower flange thickness.

y2

Y4

extreme fibre of extreme fibre of extreme fibre of extreme fibre of


Beam section (A-A) with web Saa = Beam section with web. Iaa = Beam inertia with web. y1 = distance between extreme fibre of upper flange and neutral axis. y2 = distance between extreme fibre of lower flange and neutral axis.

Study Of Flanges

Beam section (A-A) neglecting web Ssa = Beam section without web. Isa = Beam inertia without web. y3 = distance between extreme fibre of upper flange and neutral axis. y4 = distance between extreme fibre of lower flange and neutral axis.

page V1-4•6–1

MCS V1-4 • THIN WEB BEAMS V1-4 6.1.2 loads and stresses due to primary bending of beam 1) Bending moment Introducing a load T at the end of a thin web beam implies the existence of a bending moment changing along the length of the beam. The architecture of each beam span must be taken into account to evaluate the exact location where the effect of the bending moment on the flanges is the most significant. Thus, section A-A of the thin web beam shown on the previous page is the section farthest from the load T introduction point before the end fittings start to take the load off the flanges. According to the incomplete diagonal tension theory, a part of the total shear load (and therefore of the bending moment) acts in the form of pure shear, the other part in the form of pure diagonal tension. Let Mf be the total bending moment at the section under study: Mf = T * l = Mf1 + Mf2 Mf1 the bending moment concerning the beam stable in shear. Mf2 the bending moment concerning the beam in the form of pure diagonal tension. Mf1 = (1 - k) Mf and Mf2 = k Mf where k: diagonal tension factor 2) Bending stresses on extreme fibres • The bending stress on the extreme fibre of the upper flange is: M y M y σfss = − f 1 1 − f 2 3 T > 0 implies that the upper flange is compressed and Mf > 0 Iaa Isa • The bending stress on the extreme fibre of the lower flange is: Mf 1y2 M y σfsi = + f2 4 Iaa Isa 3) Normal loads in the flanges due to bending Whilst the web is stable in shear, the normal load in a flange can be expressed by: æI ö M (where h2 is the distance between the centres of gravity of each flange) F1 = çç sa ÷÷ ∗ f 1 h2 è I aa ø In the pure diagonal tension state, the normal load in the flange can be expressed by: F2 =

Mf 2 h2

Total normal loads in the flanges: Upper flange: Fssm = - F1 - F2 Lower flange: Fsim = F1 + F2. 4) Mean stresses due to primary bending in flanges F F Upper flange: σssm = ssm Lower flange: σsim = sim Sss Ssi

V1-4 6.1.3 additional loads and stresses due to the action of the tension field, mean strain The flanges are submitted to a compression load induced by the diagonal tension field: T Fc = - TD * cot α (Fc < 0) where: 2 and α is the diagonal tension angle measured in relation to the flanges. TTD = kT The stresses generated are: Upper flange: σ ssc =

Fc Sss

Lower flange: σsic =

Fc Ssi

= The mean strain to be taken into account to calculate angle α is: ε TD x


Study Of Flanges

1 2

é σ ssc σ ù + sic ú ê E si û ë E ss

Issue 0

MCS V1-4 • THIN WEB BEAMS V1-4 6.1.4 total normal loads and mean stresses in flanges Upper flange: total load Fsst = Fssm + Fc

mean stress σ sst =

Fsst = σ ssm + σ ssc Sss

Lower flange: total load Fsit = Fsim + Fc

mean stress σ sit =

Fsit = σ sim + σ ssc Ssi

V1-4 6.1.5 secondary bending moment in flanges As the web is subjected to diagonal tension, it pulls on the elements comprising the flanges; in other terms, each flange acts as a continuous beam with the uprights as supports. The transverse load on a flange is equal to the vertical flow in the web due to the diagonal tension. 1 The maximum secondary bending moment in the flanges is: Mfs = k τ e b2 Cfs 12 1

æ ö4 e ÷ Cfs: attenuation factor given on the curve below versus parameter: W d = 0.7 b çç ÷ è (I ss + Isi ) h 2 ø Remark 1: The secondary bending moment is maximum at the uprights. If k ≈ Cfs ≈ 1, the moment in the centre between the bearing points is equal to half the maximum moment with opposite sign. æ 2 W ame ö çç1 + ÷ in all cases. 3 W sem ÷ø è W sem = static moment of the flange. W ame = static moment of the efficient section of the web above the neutral fibre (for an upper flange). T τ = Sufficient approximation when flange height is low compared with the height of the beam h 2e and the flanges consist of thin sections.

Remark 2: τ =

TW sem Ie

for calculating I and W, efficient web thickness is taken as being equal to (1-k) e

Cfs versus Wd 1

Cfs

0.95 0.9 0.85 0.8 0.75 0

1

2 Wd

3

4

An approximation of Cfs versus W d is given by the following polynomial: Cfs = 0.0058W d3 - 0.0676 W d2 + 0.1252W d + 0.9366


Study Of Flanges

page V1-4•6–3

MCS V1-4 • THIN WEB BEAMS V1-4 6.1.6 stresses due to secondary bending moment in flanges UPPER FLANGE at uprights: M v σ fs 2 = − fs 2 I ss σ fs1 =

M fs v1 Iss

stress in lower fibre of upper flange stress in upper fibre of upper flange

When the upper flange bends on the bearing point, the lower fibre is compressed and the upper fibre stretched. in the centre between two uprights: In this case, M’fs is calculated in a conservative manner considering k = Cfs = 1 M' fs v 2 σ′fs 2 = stress in lower fibre of upper flange 2 * I ss σ′fs1 = −

M' fs v1 2 * I ss

stress in upper fibre of upper flange

LOWER FLANGE at uprights: M fs v 4 σ fs 4 = I si

stress in lower fibre of lower flange

M fs v 3 stress in upper fibre of lower flange I si When the lower flange bends on the bearing point, the lower fibre is stretched and the upper fibre compressed. σ fs 3 = −

in the centre between two uprights: (same remark as made above for M’fs) M' fs v 4 σ′fs 4 = − stress in lower fibre of lower flange 2 * I si σ′fs3 =

M' fs v 3 2 * I si

stress in upper fibre of lower flange

V1-4 6.1.7 overall stresses in flanges UPPER FLANGE σssTi = σssm + σssc + σfs2

overall stress, lower fibre of upper flange at uprights

σssTs = σfss + σssc + σfs1

overall stress, upper fibre of upper flange at uprights

σssTi = σssm + σssc + σ' fs2

overall stress, upper flange lower fibre in centre between two uprights

σssTs = σfss + σssc + σ' fs1

overall stress, upper flange upper fibre in centre between two uprights

LOWER FLANGE σsiTi = σfsi + σsic + σfs4

overall stress, lower fibre of lower flange at uprights

σsiTs = σsim + σsic + σfs3

overall stress, upper fibre of lower flange at uprights

σsiTi = σfsi + σsic + σ' fs4

overall stress, lower flange lower fibre in centre between two uprights

σsiTs = σsim + σsic + σ' fs3

overall stress, lower flange upper fibre in centre between two uprights


Study Of Flanges

Issue 0

MCS V1-4 • THIN WEB BEAMS V1-4 6.1.8 allowable stresses and margins The allowable stresses required to calculate the margins are as follows: * Tensile:

Strength of flanges at ultimate loads is given by the characteristics of the material.

* Compression: The instability of the flanges can be caused by:

1) under a low wavelength mode: a) Local buckling (of the flange for example) of the flanges. b) Forced crippling of the riveted section of the flanges. 2) under a medium wavelength mode: (this concerns the complete flange) a) Lateral buckling b) Column c) Crippling The details for calculating the allowable stresses are given in chapter V1-3. The margins are calculated by applying the following rule: • The compression stresses calculated taking the secondary effects into account (secondary bending or gusset effect for example) are compared with the allowable local buckling stresses (natural or forced). • The mean compression stresses in the flanges are compared with the allowable lateral buckling, column and crippling stresses. • The tensile stresses are compared with tensile breaking stress of the material.


Study Of Flanges

page V1-4•6–5

MCS V1-4 • THIN WEB BEAMS V1-4 6.2 EXAMPLE Let us take the cantilever beam of § 6.1.1 submitted at its free end to a load of 60000 N and with the following geometrical characteristics: l = 1270 mm b = 290 mm e = 0,6 mm h = 762 mm 1 h= 725,4 mm h2 = 746,4 mm 1270

60 000 N

A

0,6

290

725,4 746,4

762

A v1 ess

v2

esi

v3

Y3

y1

neutral axis

v4 neutral axis Sss = 430 mm2 Ssi = 245 mm2 Iss = 57 250 mm4 Isi = 14 780 mm4 v1 = 9,7 mm v2 = 30,3 mm v3 = 24,1 mm v4 = 5,9 mm ess = 4,5 mm esi = 2,5 mm

y2

Beam section (A-A) with web Saa = 1121 mm2 Iaa = 110,1*106 mm4 y1 = 319,4 mm y2 = 442,6 mm

Y4

Beam section (A-A) web neglected Ssa = 671 mm2 Isa = 86,1*106 mm4 y3 = 277,8 mm y4 = 484,2 mm

The mechanical characteristics of the web of the 2024 PLT3 beam are: Ea = 70300 MPa;σRa = 440 MPa;σ0,2a = 270 MPa The mechanical characteristics of the members (flanges and uprights) of the 7075 T73510 beam are: Em = 73800 MPa;σRm = 495 MPa;σ0.2m = 420 MPa The two materials have the same Poisson ratio: νe = 0.33 LIMITATIONS First of all, we must check if the beam is compatible with the limitations of § 3.2 e 4 .5 e si 2.5 • ss = = 7.5 > 0.6 and = = 4.2 > 0.6 e 0 .6 e 0.6 • 0.2 <

b 290 = = 0.38 < 1 h 762


and

h 762 120 < = = 1270 < 1500 e 0.6

Study Of Flanges

Issue 0

MCS V1-4 • THIN WEB BEAMS CRITICAL SHEAR STRESS According to § 4, the critical shear stress is given by the following formula: 2

2

é ìh b üù with ks = 3.8 * êmin í 1 ; ýú + 5.35 î b h1 þúû ëê 2 2 é ù π2 æ 290 ö æ 0.6 ö τcr, a = η + = η * 1.7 MPa 5 . 35 * 70300 * * 3 . 8 * ÷ ÷ ç ç ê ú 12 * 1 − 0.332 ëê è 725.4 ø è 290 ø ûú

τcr, a = η

æ ö π2 e ÷÷ k s E a çç min { h , b } 12 * 1 − ν e2 1 è ø

(

)

(

)

Obviously, there is no need to calculate the plasticity correction.

τcr, a = 1.7 MPa LOADING RATIO AND DIAGONAL TENSION FACTOR The loading ratio Rs and the diagonal tension factor k are defined in § 3.3 R 0.4343 − 1 τ Rs = and k = 0s .4343 τcr , a Rs +1 The flanges consist of tee sections the heights of which are low when compared with the beam. Hence, according to § 5.1.5: T 60000 τ = = = 134 MPa h2 e 746.4 * 0.6 Rs =

τ τcra

=

134 = 79 1 .7

and

k =

790.4343 − 1 = 0.74 790.4343 + 1

BENDING MOMENTS The maximum bending moment is located at section A-A (refer to § 6.1.2). 6 mm.N Mf = T * l = 60000 * 1270 = 76.2 * 10

Mf = Mf1 + Mf2 Mf1 is the bending moment concerning the beam resisting shear. Mf2 is the bending moment concerning the beam in the form of pure diagonal tension. Mf1 = (1 - k) Mf = (1 - 0.74) * 76.2 * 106 = 19.8*106 mm.N Mf2 = k Mf = 0.74 * 76.2*106 = 56.4*106 mm.N BENDING STRESSES ON THE EXTREME FIBRES OF THE FLANGES Upper flange The direction of the load T implies that the upper flange is compressed: M * y3 M * y1 19.8 * 10 6 * 319.4 56.4 * 10 6 * 277.8 σ fss = − f 1 − f2 = − − = − 239 MPa I aa I sa 110.1 * 10 6 86.1 * 10 6 Lower flange The lower flange is under tension:

σ fsi = +

M f1 * y2 M * y4 19.8 * 10 6 * 442.6 56.4 * 10 6 * 484.2 + f2 = + + = 397 MPa I aa Isa 110.1 * 10 6 86.1 * 10 6


Study Of Flanges

page V1-4•6–7

MCS V1-4 • THIN WEB BEAMS NORMAL MEAN STRESSES IN THE FLANGES DUE TO BENDING Upper flange Fssm = - F1 - F2 where: æ Isa ö M f 1 Normal load in a flange of a beam stable in shear. F1 = çç ÷÷ * h2 è I aa ø Mf 2 F2 = Normal load in a flange of a beam in pure diagonal tension condition. h2

æ 86.1 * 10 6 F1 = çç 6 è 110.1 * 10

ö 19.8 * 10 6 ÷* = 20745 N and ÷ 746.4 ø

F2 =

56.4 * 10 6 = 75563 N 746.4

The normal load in the upper flange is a compression load: Fssm = - F1 - F2 = - 20745 - 75563 = - 96308 N The mean stress in the upper flange is: − 96308 F = - 224 MPa σ ssm = ssm = Sss 430 Lower flange The normal load in the lower flange is a tensile load: Fsim = F1 + F2 = 20745 + 75563 = 96308 N The mean stress in the lower flange is: F 96308 σ sim = sim = = 393 MPa Ssi 245 STRESSES DUE TO ADDITIONAL LOADS GENERATED BY THE DIAGONAL TENSION The upper and lower flanges are submitted to a compression load induced by the shear load under the action of the tension field: T Fc = − TD * cot α where TTD = kT 2

α is the diagonal tension angle the calculation of which is detailed in chapter V1.1 and § 5.3.2. For this calculation, we will suppose that α = 45° kT 0.74 * 60000 Fc = − = − = − 22200 N 2 2 Upper flange − 22200 F σssc = c = = − 52 MPa Sss 430 Lower flange − 22200 F σsic = c = = − 91 MPa Ssi 245


Study Of Flanges

Issue 0

MCS V1-4 • THIN WEB BEAMS STRESSES DUE TO THE SECONDARY BENDING MOMENT IN THE FLANGES Secondary bending moment in the flanges (refer to § 6.1.5): 1 M fs = k τ e b 2 Cfs where Cfs is the attenuation factor given on the curve on page V1-4.6.3 versus W d 12 1

1

é ù4 é ù4 e 0 .6 W d = 0.7 * b * ê ú = 0.7 * 290 * ê ú = 2.1 Þ Cfs = 0.955 ë (57250 + 14780 ) * 746.4 û ë (I ss + I si ) h 2 û 1 Mfs = * 0.74 * 134 * 0.6 * 2902 * 0.955 = 398200 mm.N 12 Upper flange at uprights M * v2 398200 * 30.3 σ fs 2 = − fs = − = − 211 MPa Iss 57250

σ fs1 =

M fs * v1 398200 * 9.7 = = 67 MPa I ss 57250

compression stress in lower fibre. tensile stress in upper fibre.

Upper flange in the centre between two uprights (k = Cfs = 1 Þ M’fs = 563470 mm.N) M 'fs * v 2 563470 * 30.3 ′ σ fs 2 = = = 149 MPa tensile stress in lower fibre. 2 * I ss 57250 * 2

σ′fs1 = −

M 'fs * v1 2 * I ss

= −

563470 * 9.7 = - 48 MPa 57250 * 2

Lower flange at uprights M fs * v 4 398200 * 5.9 σ fs 4 = = = 159 MPa I si 14780

σ fs 3 = −

compression stress in upper fibre.

tensile stress in lower fibre.

M fs * v 3 398200 * 24.1 = − = - 649 MPa compression stress in upper fibre. I si 14780

Lower flange in the centre between two uprights (k = Cfs = 1 Þ Mfs = 563470 mm.N) M 'fs * v 4 563470 * 5.9 σ′fs 4 = − = − = − 112 MPa compression stress in lower fibre. 2 * I si 14780 * 2

σ′fs3 =

M 'fs * v 3 2 * Isi

=

563470 * 24.1 = 459 MPa 14780 * 2

tensile stress in upper fibre.

OVERALL STRESSES IN FLANGES Upper flange at uprights σssTi = σssm + σssc + σfs2 = - 224 - 52 - 211 = - 474 MPa overall stress in lower fibre. σssTs = σfss + σssc + σfs1 = - 239 - 52 + 67 = - 224 MPa overall stress in upper fibre. Upper flange in the centre between two uprights σssTi = σssm + σssc + σ' fs2 = - 224 - 52 + 149 = - 127 MPa overall stress in lower fibre. σssTs = σfss + σssc + σ' fs1 = - 239 - 52 - 48 = - 339 MPa overall stress in upper fibre. Lower flange at uprights σsiTi = σfsi + σsic + σfs4 = 397 - 91 + 159 = 465 MPa overall stress in lower fibre. σsiTs = σsim + σsic + σfs3 = 393 - 91 - 649 = - 347 MPa overall stress in upper fibre. Lower flange in the centre between two uprights σsiTi = σfsi + σsic + σ' fs4 = 397 - 91 - 112 = 194 MPa overall stress in lower fibre. σsiTs = σsim + σsic + σ' fs3 = 393 - 91 + 459 = 761 MPa overall stress in upper fibre.


Study Of Flanges

page V1-4•6–9

MCS V1-4 • THIN WEB BEAMS SUMMARY Overall stresses without secondary bending in flanges (in MPa) Flange Lower Upper upper fibre 302 - 291 lower fibre 306 - 276 Overall stresses with secondary bending in flanges at uprights (in MPa) Flange Lower Upper upper fibre - 347 - 224 lower fibre 465 - 474 Overall stresses with secondary bending in flanges at centre between two uprights (in MPa) Flange Lower Upper upper fibre 761 - 339 lower fibre 194 - 127

V1-4 6.3 CONCLUSION The calculation above shows that by taking secondary bending into account, the stresses in the flanges are significantly modified. Thus, in the centre between two uprights, the lower fibre of the lower flange changes from a tensile stress of 302 MPa to a stress of 762 MPa greatly exceeding the breaking stress of the flange (σR= 495 MPa). Whereas, at the uprights, the lower fibre of the upper flange drops from - 276 MPa to - 474 MPa exceeding the yield strength of the material of the flange (- 420 MPa). A new flange design is therefore required. Modifications can be made without adding weight to the existing flange. For instance, by extending the height of the vertical section of the flange and by adding a lip, which requires a minute reduction of the thicknesses to conserve same weight, we obtain a considerable gain for the flange quadratic moment; this being necessary to reduce the stresses due to the action of the secondary bending moment.

Flange studied in the example

Flange with same surface area but with extended vertical section. An added lip. Slightly reduced thicknesses.

In the example dealt with the web was relatively thin; also, the value of 0.74 for the diagonal tension factor k means that the folding of the web is fairly severe as 74% of the shear load is taken by the diagonal tension. For heavily loaded and fairly low beams, such as wing spars and bulkheads submitted to high external loads, the webs are much thicker and factor k much lower. Generally speaking, to avoid the forming of permanent folds, no folding of the web is accepted before 60% of the limit loads have been applied (at present, the target is 80%). Web weight saving, in relation to that which is required for a web which must not buckle, could exceed the additional weight in the web flanges and stiffeners induced by the diagonal tension field. However, the calculations for the rivets in the semi-tension field case poses more problems as the loads in the rivets are higher and more complex than in a stable-web beam.


Study Of Flanges

Issue 0

MCS V1-4 • THIN WEB BEAMS V1-4 7 END SPANS AND SPANS WITH OPENINGS V1-4 7.1 GENERAL

T

Load applied

Flange

Upright

Flange

Tension exerted by the tension field

End upright

Figure a

Opening in web

Opening framework

Figure b

The "end spans" adjacent to the folded area and the edge elements of an opening comprise a specific case. The diagonal tension pulls the end upright and the edges of an opening towards the interior; this induces secondary bending the same as the one acting on the flanges in addition to the normal compression. Figures "a" and "b" above clearly show this action. The end uprights and the edges of the opening must be much more rigid than the uprights in the standard area or at least reinforced by additional elements to reduce the stresses induced by bending.

V1-4 7.2 EDGE ELEMENTS The linear load component inducing bending in the edge elements is given by the following formulas: fby = k τ e tg α

for edge elements parallel to the neutral fibre of the beam (flanges).

fbx = k τ e cot α for edge elements perpendicular to the neutral fibre of the beam (uprights).

The greater the distance between the supports of an edge element submitted to a linear load f, the greater the bending moment that it must support. A combination of the three methods proposed below would be a good solution to relieve these elements:

• Simply reinforce the element so that it can take all the loads submitted to it. • Increase the thickness of the web in the edge area either to prevent buckling or to reduce value k and thus the linear load which introduces the bending into the edge element. • Add stiffening elements to support the edge element and thus reduce its bending moment induced by fb. The calculation for the fold angle α is detailed in §5.3.2.


End Spans And Spans With Openings

page V1-4•7–1

MCS V1-4 • THIN WEB BEAMS V1-4 7.3 EXAMPLE Let us again take the cantilever beam example of § 5.1.8 with same loads and same geometrical characteristics.

1270

60 000 N

A

0,6

290

fb

762

725,4 746,4

A End upright

The linear load inducing bending in the end upright is: fbx = k t e cot α = 0.74 * 134 * 0,6 * 1 = 59.5 N/mm (We suppose thatα = 45°) Let us consider the adding of three equally-spaced horizontal stiffeners as method for reducing the influence of the bending on the end upright and analyse the inner loads in the end span.

End span 60 000 N

290 186,6 186,6 fb 186,6 186,6

upp. flange CG

Added stiffeners

746.4 762

low flange CG End upright

The figure above shows the end span reinforced by three stiffeners spaced 186,6 mm apart. Two loads are applied to it: • The shear load at beam end: 60 000 N • The linear load component due to the tension field: fb = 59.5 N/mm. Below, we will show what effect adding three stiffeners has on the local shear in the span. The effect of each load applied will thus be studied then summed.



Issue 0

MCS V1-4 • THIN WEB BEAMS 60 000 N

60 000 N

290

23312 N

22205 N

290

45517 N

f = 57.4N/mm

f = 80.4N/mm

746,4

60 000 N

+

f = 19.1N/mm

59.5 N/mm

f = 137.8N/mm

+ 59.5 N/mm

f = 19.1N/mm

f = 99.5N/mm

f = 61.3N/mm

60 000 N f = 57.4N/mm

23312 N

f = 23N/mm

22205 N 1107 N Figure "a"

Figure "b"

Figure "c"

1) Figure "a" shows the shear flows taken as constant in each of the end span pockets and due to the load applied F = 60 000 N. Equilibrium is shown directly on figure "a" above: 60 000 * 290 The reaction at each of the flanges is : R 1 = = 23 312 N 746,4 F R1 The shear flow inside each pocket is: f1 = = = 80.4 N/mm 746,4 290 2) Figure "b" shows the mean shear flows in the centre of each of the pockets of the end span and due to the linear load component generated by the tension field: fb = 59.5 N/mm. The equilibrium is shown directly on figure "b" above: 76.6 The reaction at each flange is: 57.4 upper 59.5 * 746,4 stiffener R2 = = 22 205 N 2 19.1 centre In actual fact, the shear flow in the end span varies from 0 N/mm in 0 stiffener the centre of the beam up to a maximum at the centres of gravity of 19.1 lower the flanges: stiffener 57.4 59.5 * 746,4 f max = = 76.6 N/mm 76.6 2 * 290 3) Figure "c" shows the total shear flows obtained in the centre of each pocket of the end span and due to all loads. CONCLUSION We can see that the shear flows in the two upper pockets of the end span increase significantly above the nominal value (80.4 N/mm). This means that the effect of the diagonal tension in this area (the pocket is smaller) must be locally taken into account for checking the upper flange (in our example), the end upright, the added stiffeners, the rivets and the web, etc. As the thin web beam calculation principle supposes that the shear flow is identical in adjacent spans, the justification of the upper stiffener by the method detailed in chapter V1.1 requires the use of mean flux f and factor k values on either side of the stiffener as explained in the characteristics of a super-stiffener made symmetrical on page V1-4.2-2 (ditto for LH upright).



page V1-4•7–3

MCS V1-4 • THIN WEB BEAMS V1-4 7.4 ADDITIONAL STUDIES FOR OPENINGS IN A THIN WEB BEAM The presence of any openings in a thin web beam creates overstresses in the sheet around the holes whereas secondary bending loads may appear in the flanges and uprights. Let us consider a beam submitted to a shear load and a bending moment variable spanwise and including a rectangular opening on its horizontal axis. A

vss

C

v1

Flange CG lines

N

N

neutral fibre

h2 v2

B

vsi q

D

T Mf (CD)

Mf (AB)

Mf

Tmean

FLANGE CALCULATION 1) The primary bending causes the following stresses in the flanges at section AB: M (AB) * V1 M (AB) * V2 σf1 = f (Upper flange) and σf1 = f (Lower flange) I(AB) I(AB)

where Mf(AB): primary bending moment at section AB. V1,2: offset between extreme fibre of the upper or lower flange and the neutral fibre of the beam at AB. I(AB): inertia of beam section (section AB), opening (not necessarily rectangular) subtracted. 2) To pass along the opening, the shear load creates an additional bending moment in the flanges which is calculated by comparing these with elementary beams of length q/2 clamped at one end and free at the other and submitted to a load at end equal to the mean shear load in the opening area.

T q/2 q upright

upright

if no load distributed between the uprights

T/2 T/2

upright

The breakdown of the moment induced by the mean shear load of the span (no load distributed between the uprights) prorata the upper and lower flange inertias (Iss and Isi ) gives: I ss q I si q M fss = *T* and M fsi = * T * . The resulting stresses are: I ss + I si 2 I ss + I si 2 M fssi * v si T * q * v ss T * q * v si = and σ fsi = Iss 2 * (I ss + I si ) I si 2 * (I ss + I si ) v = dimension between the extreme fibre of the flange (upper or lower) and its neutral fibre. For a variable-height beam, refer to chapter V2.4.

σ fss =

M fss * v ss

=

3) Sum total stresses at section AB:



Issue 0

MCS V1-4 • THIN WEB BEAMS upper flange side: σss = σf1 + σfss


lower flange side: σsi = σf2 + σfsi


page V1-4•7–5

MCS V1-4 • THIN WEB BEAMS V1-4 8 STUDY OF WEB V1-4 8.1 STRESSES IN WEB V1-4 8.1.1 In pocket thicknesses Refer to V1-1.5.7.1. The pocket thicknesses, assumed to be identical on either side of the upright are subjected to plane stresses. k τ tan α

τ TDP

Tresca criterion:

τ max = τ

2

πö æτ æ ö æ k ö 1+ ç ÷ ; ç α = ÷ Þ ç max = 1÷ 4ø è τ è è tan 2α ø ø

V1-4 8.1.2 At uprights Refer to V1-1.5.7.2. In shaded section at web - upright fastener row:

et Tresca criterion:

τ max e = 1,3 τ et

æ k ö 1+ç ÷ è1 + k ø

2

e

V1-4 8.2 WEB FAILURE Refer to V1-1.5.9.4. Generally, the skin is thicker at the uprights than in the pockets. Failure can then occur either in the pocket-land blendin radii or at the web-stiffener fastener rows. • Failure in pocket thickness (TRESCA): τmax adm = • Failure at web - upright fasteners: τ' max adm =


σR 2

σR 2

Study Of Web

page V1-4•8–1




Study Of Web

Issue 0

MCS V1-4 • THIN WEB BEAMS V1-4 9 STUDY OF FASTENERS V1-4 9.1 WEB-FLANGE AND WEB-END UPRIGHT CONNECTIONS MINIMUM SHEAR STRENGTH Upright

Upright

Upright Flange

y

A

Lt

Lt

A-2Lt

Flange x The stress is not constant along a pocket: • As the upright is compressed, a portion of the web 2 Lt long is submitted to compression. • The centre of the pocket is submitted to diagonal tension as shown above over length A-2Lt. Thus: * The flows acting on the fastener rows connecting the flange to the web are: ty nx 2k = cot α ; = 1 τe τe 1 +k These flows are supported by any line of fasteners parallel to the flanges. For stiffened panels: junction parallel to crossmembers. * The flows acting on the rows of fasteners connecting the end upright to the web are: ny 2k t = tan α ; x = 1 τe τe 1 +k For stiffened panels: junction parallel to stiffeners. End upright Flange

ps

Row of fasteners

y ty nx

pm tx ny

x


Study Of Fasteners

page V1-4•9–1

MCS V1-4 • THIN WEB BEAMS TENSILE STRENGTH The fasteners must also have a sufficient tensile strength to prevent the buckled sheet from separating from the flanges. The required tensile strength for a fastener is given by the following criterion: F(fastener failure) ≥ 0.15 * σR * p * e

where:

σR: web tensile breaking stress p: e:

rivet pitch web thickness

V1-4 9.2 STANDARD DISSYMMETRICAL WEB-UPRIGHT CONNECTIONS (SINGLE ANGLE) SHEAR STRENGTH No criteria concerning the shear strength of the fasteners on a single angle has been established by NACA. The continuity of the web on either side of the upright means that the standard fasteners are subjected to no loads (except in cases of separation between the web - upright elements). However, at the ends of the uprights, the fasteners are submitted to loads. The tensile strength criterion is probably sufficient to ensure a satisfactory design. TENSILE STRENGTH The required tensile strength for a fastener is given by the experimental criterion: F(fastener failure) ≥ 0.22*σ σR*p*e

(The tensile strength of a fastener is defined as being the tensile load which will cause any failure; if the sheet is thin, the failure will consist of the fastener being pulled through the sheet). Spacing between fasteners on single angles must be sufficiently low to prevent inter-rivet buckling (or buckling of the flange of the angle if its thickness is lower than that of the web), at a compression stress level equal to σ TD . This y min spacing must also be lower than b/4 to justify the supported edge assumption made when determining the critical web stress.

V1-4 9.3 UPRIGHT-WEB CONNECTIONS These fasteners must take the load applied by the web and transferred to the upright. The loads are as follows: Pm = σ TD y * Sm Pm = σ TD * y

for a double angle. Sm

ægö 1 + çç ÷÷ èρø

2

for a single angle.


Study Of Fasteners

Issue 0

MCS V1-4 • THIN WEB BEAMS V1-4 10 EXAMPLE Let us again take the cantilever beam of §5.1.1 submitted at its free end to a load of 60000 N and with the following geometrical characteristics: l = 1270 mm b = 290 mm e = 0,6 mm h = 762 mm 1 h= 725,4 mm h2 = 746,4 mm

60 000N

0,6

290

762

725,4 746,4

3,2 v1 v2

ess

esi

v3

25 7,4

v4 Upper flange

Sss = 430 mm2 Iss = 57 250 mm4 ess = 4,5 mm v1 = 9,7 mm v2 = 30,3 mm

Lower flange Ssi = 241 mm2 Isi = 14 780 mm4 esi = 2,5 mm v3 = 24,1 mm v4 = 5,9 mm

25 Upright (single angle) Sm = 150 mm2 Im = 8654 mm4 em = 3,2 mm

The mechanical characteristics of the web of the 2024 PLT3 beam are: Ea = 70300 MPa;σRa = 440 MPa;σ0.2a = 270 MPa The mechanical characteristics of the members (flanges and uprights) of the 7075 T73510 beam are: Em = 73800 MPa;σRm = 495 MPa;σ0.2m = 420 MPa The two materials have the same Poisson ratio: νe = 0.33 REMINDER The limitations, the critical shear stress, the loading ratio and the diagonal tension factor were checked in § 5.1.8: e 3,2 From among the limitations, we must check: m = = 5,3 > 0,6 e 0,6 τcr,a = 1.7 MPa We recall that: τ = 134 MPa Rs = 79 k = 0.74


Example

page V1-4•10–1

MCS V1-4 • THIN WEB BEAMS CALCULATING UPRIGHT CHARACTERISTICS 290 Lt Load-carrying width: Lt TD = (1 − k ) 0 = (1 − 0,74 ) = 19 mm 4 2 Load-carrying section: 290 ∆STD = (1 + k) Lt0e = (1 + 0,74) * * 0,6 = 151,2 mm2 y 2 2 S0 = Sm + 2Lt0e = 150 + 290 * 0,6 = 324 mm 2 TD TD S y = S0 - ∆S y = 324 - 151,2 = 172,8 mm

X

X LtTD

LtTD

Web characteristics: 2 Sa0 = b * e = 290 * 0,6 = 174 mm 4 IXXa0 = 179 mm W XXa0 = - 174 mm3 da0 = - 1 mm e 3 TD W XXa = W XXa0 + ∆STD y = - 174 + 0,3 * 151,2 = - 128,6 mm 2 æ e 2 e 2 ö TD æ 0,6 2 0,6 2 ö ç ÷ çç ÷÷ * 151,2 = 161 mm 4 I TD = I + ∆ = − + S 179 XXa0 XXa ç 4 12 ÷ y 4 12 è ø è ø Characteristics of the load-carrying section of the upright after correction due to the various materials æ STD ö æ Sm ö ÷ E = æ 22.8 ö * 70300 + æç 150 ö÷ * 73800 = 73338 MPa Ey = ç aTD ÷ E a + ç TD ç Sy ÷ ç S y ÷ m çè 172.8 ÷ø è 172.8 ø è ø è ø æ E ö TD æ E m ö æ 70300 ö æ 73800 ö TD 3 ÷W W XX = ç a ÷ W XXa +ç = * − 128.6 + ç ÷ * 1111.6 = 995 mm ç Ey ÷ ç E y ÷ XXm çè 73338 ÷ø 73338 ø è è ø è ø

dTD =

TD W XX 995 = 5,76 mm = TD 172.8 Sy

æ Ea ç I TD XX = ç Ey è

ö TD æ ö ÷ I XXa + ç E m ÷ I XXm = æç 70300 ö÷ * 161 + æç 73800 ö÷ * 8654 = 8863 mm 4 ÷ ç Ey ÷ è 73338 ø è 73338 ø ø è ø 2

TD TD ITD = I TD = 8863 - 172.8 * 5.762 = 3130 mm4 XX − S y d

Characteristics of the load-carrying section of the lower flange (less rigid) after correction due to the various materials h1e 725,4 * 0,6 ∆STD = (1 + 0.74) * = 189 mm 2 x = (1 + k ) 4 4 h1e ö 725,4 * 0,6 ö æ æ TD 2 STD ÷ − ∆SX = ç 241 + ÷ − 189 = 269,6 mm X = ç Ssi + 2 ø 2 è è ø 2 TD STD Xa = S X − Ssi = 269,6 - 241 = 28,6 mm

æ STD ö Ssi ÷ * E a + TD Ex = çç Xa * Em = TD ÷ SX è SX ø


æ 28,6 ö æ 241 ö çç ÷÷ * 70300 + çç ÷÷ * 73800 = 73429 MPa è 269,6 ø è 269,6 ø

Example

Issue 0

MCS V1-4 • THIN WEB BEAMS CALCULATING THE STRESSES IN THE UPRIGHT (WITH ASSOCIATED WEB) Web orientation compression:

k Lt 0 e τ 725,4 = − 0.74 * * 0.6 * 134 * cot α = - 21579 cot α tan α 2

N TD = − x σ TD = x

N TD − 21579 cot α x = = − 80 cot α TD 269,6 Sx

ε TD = x

σ TD − 80 cot α x = = − 0.109%. cot α Ex 73429

Upright orientation compression: N TD y = - k b e τ tan α = - 0.74 * 290 * 0,6 * 134 * tan α = - 17254 tan α

σ TD = y εTD = y

N TD y S

TD y

σTD y Ey

= =

− 17254 tan α = − 99.85 tan α 172,8 − 99.85 tan α = − 0.136% tan α 73338

Diagonal tension:

As the fold angle is around 45°, we assume sin 2α = 1 τ 134 ε TD ≈ [1 + k + ν e (1 − k )] = [1 + 0,74 + 0,33(1 − 0,74)] = 0,348% Ea 70300 Diagonal tension angle: æ ö ε − ε TD 0.348 + 0.109 X (α0 = 45°) → ç tan 2 α1 = TD = = 0.944 Þ tan α1 = 0.972 Þ α1 = 44.18° ÷ TD ç ÷ ε TD − ε y 0.348 + 0.136 è ø æ ö ε − ε TD 0.348 + 0.109 / 0.972 X (α1 = 44.18°) → ç tan 2 α 2 = TD = = 0.958 Þ tan α 2 = 0.979 Þ α 2 = 44.39° ÷ TD ç ÷ ε TD − ε y 0.348 + 0.136 * 0.972 è ø TD æ ö ε − εX 0.348 + 0.109 / 0.979 (α2 = 44.39°) → ç tan 2 α 3 = TD = = 0.955 Þ tan α 3 = 0.977 Þ α 3 = 44.34° ÷ TD ç ÷ ε TD − ε y 0.348 + 0.136 * 0.979 è ø TD æ ö ε − εX 0.348 + 0.109 / 0.977 (α3 = 44.34°) → ç tan 2 α 4 = TD = = 0.956 Þ tan α 4 = 0.978 Þ α 4 = 44.35° ÷ TD ç ÷ ε TD − ε y 0.348 + 0.136 * 0.977 è ø finally: α = 44.35°; tanα = 0.978 Mean and minimum compression stresses in the upright. σ TD = - 99.85 tanα = - 99.85 * 0.978 = - 97.7 MPa y

σ TD y min σ

TD y

æ b ö ÷ + k = (1 − 0.74) = (1 − k ) çç1.78 − 0.64 * h m ÷ø è

290 ö æ çç1.78 − 0.64 * ÷ + 0.74 = 1.14 725 ,4 ÷ø è

TD σ TD = - 111.4 MPa y min = 1.14 * σ y

REMARKS: The upright orientation and flange orientation stiffening coefficients are close to each other: 324 − 290 * 0,6 466 − 375 * 0,6 upright orientation coefficient: = 0.46; flange orientation coefficient: = 0.52 324 466 We were justified to use the following approximation (difference negligible in relation to the solution above). (α ≈ 45°) Þ ( σ TD = - 99.9 MPa;σ TD y y min = 1.14 * - 99.85 = - 113.9 MPa)


Example

page V1-4•10–3

MCS V1-4 • THIN WEB BEAMS STRESSES IN WEB ALONE AND IN UPRIGHT ALONE Shear in web (refer to V3-1 § 5.7.1)

(α ≈ 45°) Þ (τmax = τ = 134 MPa) Shear in web at rivet row (refer to V3-1 § 5.7.2) e τmax = τmax = τ = 134 MPa τ' max = et Compression stress in upright orientation web Ea 70300 TD σ TD = − 111.4 * = − 106.8 MPa ya min = σ ymin * Ey 73338 Compression stress in upright alone E 73800 σ TD = σ TD * m = − 97.7 * = − 98.3 MPa m y Ey 73338 TD σ TD m min = σ y min *

Em 73800 = − 111.4 * = − 112.1 MPa Ey 73338

ALLOWABLE STRESSES The upright consists of a single angle; we will calculate below the allowable column buckling and Forced crippling stresses. Column buckling stress

• No stress in the upright exceeds the elastic stress of the material: 420 MPa. • Calculation of allowable stress at zero slenderness ratio supposing, to limit calculations, that the limitation comes from the web (refer to V3-1 § 4.6.4): σ0a = 270 MPa;ε0 =

270 + 0.002 = 0.584% 70300 13.83

σ0m = 385.6 MPa because:

0.584 385.6 æ 385.6 ö = + 0.002 * ç ÷ 100 73800 è 420 ø

æ STD ö æ Sm ö ÷ σ = æç 22.8 ö÷ * 270 + æç 150 ö÷ * 385.6 = 370 MPa σ0 = ç aTD ÷ σ 0a + ç TD ç Sy ÷ ç S y ÷ 0 m è 172.8 ø è 172.8 ø è ø è ø The mean allowable stress at zero slenderness ratio is: 370 MPa.

ρ =

I TD = STD y

3130 h 725,4 100 100 = 4,3 mm;λ = m = = 84; σ 0 * = 370 * = 0.5 172,8 2ρ 2 * 4 .3 Ey 73338

According to the diagram on page V3-1.4/23: σ * 100 σ for λ = 84 and 0 = 0.5 we obtain : crit = 0.28 Þ σcrit = 0.284 * 370 = 105 MPa Ey σ0 The critical column buckling stress is - 105 MPa. The mean stress in the straight section of the upright is – 97.7 MPa and does not exceed - 105 MPa. Forced crippling stress

(es = 3,2 mm;ae= 0,6 mm;σ0.2m = 420 MPa; mE= 73800 MPa; aE= 70300 MP; k = 0.74) 1

σ flf = − 0.051

æ e E ö3 k 3 çç s m ÷÷ = − 0.051 è ea E a ø + 0.002

σ 0.2 m σ 0.2 m Em

2

420 420 + 0.002 73800

2

1

æ 3.2 73800 ö 3 ÷÷ = − 354.8 MPa 0.74 çç è 0,6 70300 ø 3

The stress σTD m min = - 112.1 MPa does not exceed the stress of – 354.8 MPa.


Example

Issue 0

SSM V1-5 • STABLE WEB BEAMS CONTENTS

V1-5 V1-5 V1-5 V1-5 V1-5 V1-5 V1-5 V1-5

STABLE WEB BEAMS 1 GENERAL 2 STUDY OF FLANGES 3 STABLE WEBS 4 STUDY OF UPRIGHTS 5 JUNCTIONS 6 EXAMPLE 7 STABLE WEB BEAMS WITH ROUND LIGHTENING HOLES.

V1-5 1

issue

date

1 1 1 1 1 1 1 1

4/1999 4/1999 4/1999 4/1999 4/1999 4/1999 4/1999 4/1999

GENERAL

change

1–1

V1-5 1.1 DEFINITIONS

1–1

V1-5 1.2 BEHAVIOUR OF A STABLE WEB BEAM

1–2

V1-5 1.3 FAILURE MODES

1–3

V1-5 2

STUDY OF FLANGES

2–1

V1-5 2.1 ALLOWABLE STRESSES

2–1

V1-5 3

STABLE WEBS

3–1

V1-5 3.1 CRITICAL STRESSES IN WEBS

3–2

V1-5 4

STUDY OF UPRIGHTS

4–1

V1-5 5

JUNCTIONS

5–1

V1-5 5.1 GENERAL FORMULAS

5–1

V1-5 5.2 JUNCTIONS BETWEEN REINFORCING PLATES AND FLANGES

5–2

V1-5 5.3 WEB SPLICES

5–3

V1-5 5.4 EXAMPLE

5–4

V1-5 6

EXAMPLE

6–1

V1-5 7

STABLE WEB BEAMS WITH ROUND LIGHTENING HOLES

7–1

V1-5 7.1 GENERAL

7–1

V1-5 7.2 STANDARDISATION OF FLANGED HOLES

7–1

V1-5 7.3 ALLOWABLE SHEAR FLOW

7–2

Issue 1  AEROSPATIALE - 1999

Contents

page V1-5•i

SSM V1-5 • STABLE WEB BEAMS V1-5 7.4 REMARKS

7–5

V1-5 7.5 EXAMPLE

7–5

page V1-5•ii  AEROSPATIALE - 1999

Contents

Issue 1

SSM V1-5 • STABLE WEB BEAMS SYMBOLS USED a: Pocket length (between fastener rows) b: Pocket width (between fastener rows) B: Width of a flange C: Distance between the edges of 2 holes C’: C-2B d: Diameter of a rivet or of a hole D: Diameter of a hole including the stamping e: Thickness of the web E: Young’s modulus Ec: Young’s modulus in compression Ep: Young’s modulus - panel f: Shear flow fadm: Allowable shear flow fp: Shear flow in a reinforcing plate Fadm,cis: Allowable shear force FMf: Force induced by the bending moment FR: Resulting force FS: Tensile force on a flange FT: Force induced by shear force g: Distance between hole centrelines ha: Height of the web (between datum lines) H: Height of the beam or of a flange I: Bending rigidity constant (inertia) Im: Inertia required for an upright Iss: Upper flange inertia Isi: Lower flange inertia IA.N: Inertia in relation to neutral fibre I0: Inertia of an item in relation to its own centreline Ks,ks: Buckling coefficients in shear

Lt0: Initial load-carrying width Lt: Load-carrying width M: Moment Mf: Bending moment Mf,ame: Bending moment in web Mmax: Max bending moment n: Work hardening coefficient (Ramberg and Osgood), or number of rivets N: Normal tensile force p: Rivet pitch r: Distance between a rivet and the centre of the group of rivets Rf: Loading ratio in single bending Rs: Loading ratio in single shear S: Area Sc: Reduced shear section Sp: Area of reinforcing plates Ss: Total area of a flange Sss: Upper flange area Ssi: Lower flange area T: Shear force Tcr: Critical shear force v: Calculation dimension in relation to neutral fibre W: Static moment W Semelle: Static moment flange section alone y:Distance between the centreline or the neutral fibre and the cg of the item y : Offset between the centreline and the neutral fibre

Kf,kf: Buckling coefficients in bending L: Spacing between uprights (between datum lines)


Contents

page V1-5•iii

SSM V1-5 • STABLE WEB BEAMS ε: Strain (expansion)

τ: Shear stress

η: Plasticity correction coefficient

τR: Breaking shear

νe: Poisson ratio - elastic

τcr: Critical shear

σmat: Allowable bearing stress σ0.2: Conventional yield strength

τcr,0: Critical shear stress alone

σR: Breaking tensile stress

τh, τc: Collapsing stress in shear for a plate of length h or c

σcr: Critical stress

τmax: Max shear stress

σcrip: Critical crippling stress

τnet: Net shear stress at a hole

σcp: Critical panel compression stress σf: Bending stress σcf,0: Critical bending stress alone

page V1-5•iv  AEROSPATIALE - 1999

Contents

Issue 1

SSM V1-5 • STABLE WEB BEAMS BIBLIOGRAPHY

1- BRUHN ANALYSIS & DESIGN OF FLIGHT VEHICLE STRUCTURES


Contents

page V1-5•v

SSM V1-5 • STABLE WEB BEAMS


page V1-5•vi  AEROSPATIALE - 1999

Contents

Issue 1


V1-5 1 GENERAL This chapter mainly deals with the means for lightening thin profile beams without buckling the webs. The case where the web folds and becomes the seat of a diagonal tension field is studied in chapter V1-4. The general beam theory is developed in volume 2 of MTS004. The structural weights of stable web beams are not always optimal when compared with thin web beams. They must however be preferred each time that the stiffness of the beam is a design criterion. The first part of this chapter gives an exhaustive study of flat web beams equipped with uprights. The second part of this chapter deals with beams with webs drilled with flanged holes. This type of beam can provide a gain in weight when compared with the full web beam stiffened by relatively spaced uprights.

V1-5 1.1 DEFINITIONS A beam is said to be a stable web beam if while supporting the loads it conserves its initial flatness (no buckling of the web). The tangential stress remains lower than the critical buckling stress of the webs which are considered as plates bearing on their members with rigidity sufficient to prevent all global buckling of the web. We recall that the buckling stress of the web is not necessarily a failure stress. Thus, to be stable, a web is not loaded to its maximum capacity. Its stiffeners or uprights which are used only to prevent local buckling must also be stable to avoid any problems due to crushing. To obtain a satisfactory strength/weight ratio, the flanges must be designed to make the radius of gyration of the cross section of the beam as large as possible whilst themselves having a cross section resisting buckling. Also, for long cantilever beams (the spars of a cantilever wing for instance) the flange sections must be tapered, that is a reduction in the section from the root towards the tip.


General

page V1-5•1–1

SSM V1-5 • STABLE WEB BEAMS V1-5 1.2 BEHAVIOUR OF A STABLE WEB BEAM Only one bending plane is considered.

x

T

y

Mf

a

a

The external forces which can be applied to a stable web beam are: • T: Shear force • Mf: Bending moment The webs of thin web beams, which are unstable, take only the shear and the diagonal tension. Here, the behaviour is in compliance with the plane bending theory. The shear force is contained in the plane of the web and the flanges can be compared with plates in the planes almost perpendicular to that of the web. Therefore, the web takes almost alone the shear force whereas the highest normal stresses are taken by the flanges. Schematically, the web is submitted to shear and bending stresses whereas the flanges are submitted to normal tensile or compression forces. The presence of the uprights stabilises the web and limits the local strains due to point loads. Compared with a beam without uprights subjected to the same forces, the thickness of the web can be reduced thus leading to a significant gain in weight for very high beams. The uprights are dimensioned to ensure at least correct web supporting conditions.

page V1-5•1–2  AEROSPATIALE - 1999

General

Issue 1


V1-5 1.3 FAILURE MODES The causes of failure in a stable web beam are: • Flange tensile failure. • Local buckling of flanges in compression. • Failure of the web in shear. Tests are required to accurately determine the allowable stresses in the beams. However, in most cases, the material curves (in tension and in compression) and the limitations given by the buckling calculations in compression comprise a sufficient database to evaluate the allowable stresses. It often happens that the stresses calculated at the extreme fibre of a section are above the yield strength of the material which means that a portion of the section will buckle before the others and, as in certain cases, several different materials are used in the same section, a general method for calculating the ultimate strength of the section consists in using the stress-strain laws of the materials.

In a stable web beam, the following must be checked: • The stability of the web (plate submitted to the combined effects of compression, bending and shear). • The stability of the compressed flange. • The rigidity of the uprights ensuring the stability of the web. • The non-failure condition of the tensioned flange. • The strength of the attachments. • The shear strength of the web.


General

page V1-5•1–3




General

Issue 1


V1-5 2 STUDY OF FLANGES The flanges of a stable web beam are mainly submitted to normal forces generated by the general bending of the beam. The stability of the compressed flange and the non-failure of the tensioned flange must be ensured.

Tensioned flange Mf

Compressed flange

V1-5 2.1 ALLOWABLE STRESSES COMPRESSED FLANGE The allowable stresses in a flange submitted to compression are those guaranteeing good resistance to all local buckling phenomena. The calculation of these stresses was studied in chapter V1-3.

TENSIONED FLANGE The allowable stress is the failure tensile stress.


Study Of Flanges

page V1-5•2–1




Study Of Flanges

Issue 1


V1-5 3 STABLE WEBS The web, designed not to buckle under the action of the design loads is capable of supporting the plane bending stresses in addition to the shear force. In the elastic range, the stresses due to the bending moment and to the shear force are: M v • σf = f where I is the inertia of the complete cross section of the beam. I T • τ= y dS (BREDT formula). Ieò

As integral

ò y dS

is maximum at the neutral fibre =>τmax is located on the neutral fibre.

If the material is concentrated in the flanges, the shear stress in the web is almost constant and is around: T T τ= = h.e Sc Sc: Reduced shear section. I . h= W W: Static moment of the 1/2 section. Example: 10 5 G

I

100

M

50 10 G Figure (a)

50

Figure (a): Shear stress at G:

τG =

T I

=

.e

W

T 84.8 × 5

=

Figure (b)

T 424

Figure (b): Shear stress at M:

τM =

T I W

Þ

.e

=

T 99.9 × 5

=

T 499.3

τM = 0.85 τG


Stable Webs

page V1-5•3–1

SSM V1-5 • STABLE WEB BEAMS V1-5 3.1 CRITICAL STRESSES IN WEBS The critical stresses in the webs are calculated in the same way as described for a flat plate in chapter V1-2 "Buckling of plates and thin shells". The web of the beam is considered as simply supported on its edges (flanges and uprights). The uprights must provide sufficient stiffness to prevent the buckling of the pocket and also so that they can be considered as a support (see next paragraph).

T

a Mf

e

b

Let: a = distance between uprights (between fastener rows); b = height of beam (between fastener rows); e = web thickness

SHEAR (see V1-2.2.3.1.2)

τ cr,0

ö e π 2k s E æ =η ÷ 2 ç 12 1 − ν e è min{a, b}ø

(

2

RS =

)

τ τcr,0

σ1

BENDING (see V1-2.2.3.1.3 )

σ cf,0 = η

π2 k f E æ eö 2 2 ç ÷ 12 1 − ν e è bø

(

Rf =

)

σf σcf ,0

Mf

σf =

σ1 − σ 2 2

σ2 SHEAR + BENDING (see V1-2.2.4.1) R s + R f = 1 where R.F. = 2

1

2


R + R2f 2 s

−1

Stable Webs

Issue 1


V1-5 4 STUDY OF UPRIGHTS The moment of inertia to be given to a cross section of an upright intended to prevent the buckling of the web under shear is: 4

Im

2.29 L æ Tcr h a ö 3 = ç ÷ e è 33 E ø

L

Im: Moment of inertia required for the upright.

e

ha

L: Spacing between uprights (between datum lines). e: Web thickness. Tcr: Critical vertical shear force in section. ha: Web height (between datum lines). E: Young’s modulus of elasticity. As Tcr = τcr h a e

æ ö e τ cr = Ks E ç ÷ è min{L, ha }ø

and

2

hence:

τcr: Web critical shear stress. Ks: Buckling coefficient in shear. The curve below is used in the following cases: • All flat or curved panels. • The uprights are attached at both ends (flanges). • No participation of the effective skin must be included in the calculation of the moment of inertia. • The curve is only applicable to cases where buckling is elastic or where L = min {L , h a } The moment of inertia required for the upright can be written in another way:

4

Im =

2.29 L æ τ cr h a e ö 3 2

e

ç ÷ è 33 E ø

2 æ e ç Ks E æç ö÷ h 2a è Lø 2.29 L ç = ç e 33 E ç è

4

ö3 e÷ 3 ÷ = 0.0217 L e 8 ÷ æ L ö3 ÷ ÷ ç ø çh K ÷ è a sø

=>

Im Le

3

=

0.0217 8

æ L ö3 ç ÷ çh K ÷ è a s ø

When the upright is used as such and not as a means for transferring an outside load concentrated on the web of the beam, the question to be asked is what is the minimum number of rivets required to connect the upright to the web? For stable webs, we suggest two criteria: 1. The upright must be attached to the flanges at both ends. 2. The spacing between the rivets must be at most equal to a quarter of the spacing between the uprights or a quarter of the height of the web if this is smaller so as to justify the simply supported hypothesis on the edges of the web panel.


Study Of Uprights

page V1-5•4–1


Curve for calculating the minimum inertia required for the upright

1000

lm / (L * e3)

100

10

1

0,1

0

0,04

0,08

0,12

0,16

0,2

L / (ha * Ks0,5)


Study Of Uprights

Issue 1


V1-5 5 JUNCTIONS V1-5 5.1 GENERAL FORMULAS In general, standard beams consist of separate web and flange elements assembled by rivets, bolts or other means. We must know the loads that these fasteners will take to give them a suitable strength.

p C

GFlange

A

Fs + ∆ Fs

Fs

rivet D

Q

B

fxy

T

d

M+∆M

fyx

M

neutral fibre

GComplete section T

B’

D’ Fi + ∆ Fi

rivet

Fi

A’

C’

The figure above shows a beam portion equal in length to the riveting pitch "p". The section at (AA’) is submitted to a bending moment M, the section (CC’) is submitted to moment M+ ∆M. The distribution of the bending stresses in the beam sections is given by the triangular profiles assuming that the web takes no bending stresses. Let Fs be the total tensile force on the sections comprising the upper flange AB due to the bending stresses induced by M in section AA’. Also, the total tensile force on the sections comprising the flange CD due to the bending moment M+∆M is equal to Fs + ∆Fs. Under the action of these two forces, the sections comprising the upper flange must move towards the left; this movement is prevented by the rivet which attaches these angles to the web. Therefore, the force on the rivet is equal to∆Fs (same reasoning for the lower flange).

ò σ dS

Fs =

where: σ =

Semelle sup

Fs =

M I

ò x dS =

Semelle sup

M Ws I

M x I

where: Fs: force normal to upper flange.

Also: M M ò x dS = I W i I Semelle inf We deduce: ∆M Ws ∆Fs = I Fi =

where: Fi: force normal to lower flange.

and

∆Fi =

∆M Wi I

Now: ∆M = T × p

Þ

∆Fs =

T. W s

∆Fi =

T. W i

.p I I ∆Fs and ∆Fi represent the horizontal shear flow caused by the flanges over a length p.


.p

and

Junctions

page V1-5•5–1

SSM V1-5 • STABLE WEB BEAMS The fasteners ensuring the cohesion of the beam must be capable of transferring the shear flows given by T.W . formula: f = I The fasteners attaching the web to the flange are submitted to shear flow fxy, passing from the web into the flange. At point Q, the shear flows equal: T f xy = fyx = W semelle . I where: I = Moment of inertia of complete beam section. W semelle = Static moment of flange section alone. W semelle = ò y dS = Ssemelle × d We can deduce the force per fastener: ∆F =

T. W semelle I

×p

Specific case: If the two flanges are identical and the section of the web is low: 1 = 2 x Ssemelle x d2 T Þ f= 2d Approximation: By taking 2d = H the height of the beam, we obtain results close to the general expression. Þ

f=

T H

V1-5 5.2 JUNCTIONS BETWEEN REINFORCING PLATES AND FLANGES The figure below shows typical beam flanges consisting of a main (T-shaped) element reinforced by plates (a) and (b) for the upper flange and (c) and (d) for the lower flange. The aim of the fasteners is to prevent the reinforcement plates from sliding along the main (T-shaped) element of the flange by the bending action of the beam. This horizontal force which tends to make the reinforcing plates slide, this being prevented by the shear of the rivets, is given by the fundamental shear flow equation: T f = ò y dS I fa1

fa2 a b

GComplete section

ha

Hd T

hc

Shear flow

GComplete section

1st shear plane

Upper flange

c

d

2nd shear plane Lower flange

Upper flange (LH portion): Flow passing through shear plane:

f a1 =

Where: W a1 = S a1 × h a


Junctions

T. W a1 I

Issue 1

SSM V1-5 • STABLE WEB BEAMS Lower flange (LH portion): • Flow passing through 1st shear plane: Where: W

c1d1

f c1d1 =

= Sc1 . hc + Sd1 . hd

• Flow passing through 2nd shear plane: Where: W

d1

T.W c1d1 I

f d1 =

T.Wd1 I

= Sd1 . hd

The reasoning is the same for the RH portions of the plates. Where: index 1: LH portion of the plate. index 2: RH portion of the plate.

V1-5 5.3 WEB SPLICES Splices are often used in the design of a beam with thin sheet sections either to increase the thickness of the web or to ensure a junction between two beams. The case shown on figure (a) below uses splices to avoid the joggling of a relatively heavy web as case on figure (b).

a ra

Overlap junction

Splices

Figure (b)

Figure (a) x fy

fx

x2

Calculation of normal flow:

fy = σy . e

Calculation of tangential flow:

fx =

T.W (x ) I

x2

Where: W (x ) =

ò x dS

x1

y

x1

W (x) static moment of the section above x

Remark: If we consider fx as constant (conservative): Max. shear flux: f x =

T.W I


where

W = 1/2-section static moment.

Junctions

page V1-5•5–3

SSM V1-5 • STABLE WEB BEAMS V1-5 5.4 EXAMPLE Let us check the riveting of the beam below supposing that the vertical shear force on this section is equal to 15000 N. 44 38 1,6

8

Detail of flange sections

1,3

30 All rivets ASN-A 2051 diameter 3.2 mm

19 Neutral fibre 14,3 Centreline

178

5,3 19

Web height 175 mm 1,6

8

Surface area = 58 mm2 Inertia = 2020 mm4

Tensioned flange ITEM

Upper flange Upper reinforcement plate Web Lower flange Lower flange rivet hole Total

AREA (S) mm2 116

Y mm 83,7

Sy mm3 9709

Sy2 mm4 812660

I0 mm4 4040

I = I0 + S y 2 mm4 816700

70

89,8

6286

564483

negligible

564483

228 116

0 - 83,7

0 - 9709

0 812660

580600 4040

580600 816700

- 14

- 78

1092

- 85176

negligible

- 85176

516

7378

2693307

I0: moment of inertia of an element in relation to its own axis. y: distance between the centreline and the centre of gravity of the element. The offset between the centreline and the neutral fibre is given by: y = The moment of inertia in relation to the neutral fibre is: 2 2 4 I A.N = I − å S y = 2 693 307 - 516*14, 3 = 2 587 790 mm

(

å Sy = 7378 = 14, 3 mm å S 516

)

CHECKING WEB-FLANGE RIVETING T The shear flow is given by formula: f = y dS I A .N ò The beam bending load is supposed to be such that the upper flange is compressed. The first step consists in determining the moment of inertia of the cross section in relation to the neutral fibre as shown in the calculation details of the table above, by calculating the moment of inertia at the centreline of the section then transferring it to the neutral fibre. The force passing through the rivets attaching the upper flange angles to the web:

F = f * p ( p = rivet pitch = 30 mm)


Junctions

Issue 1

SSM V1-5 • STABLE WEB BEAMS The static moment of the flange in relation to the neutral fibre is:

ò ydS = 2 *58*(89 − 14, 3 − 5, 3) = 8050 mm

3

The static moment of the reinforcing plate in relation to the neutral fibre is:

ò ydS = 44 *1, 6*(89 − 14, 3 + 0,8) = 5315 mm

3

3 The total static moment of the web - upper flange junction in relation to the neutral fibre is: 13365 mm

The force in the rivet is then: é T F = f *p = ê ë IA .N

ù

15 000

ò ydSúû *p = 2 587 790 *13 365* 30 = 2324 N

The shear flow calculated by the simplified equation can be written: T f= where H is the height of the beam. H The force in the rivet is then: F = f * p =

T H

*p =

15 000 179.6

* 30 = 2506 N

This result is correct if compared with the 2324 N obtained by the exact theory. The web is attached to the two sections of the flange by 3.2 mm diameter aluminium alloy ASN-A 2051 rivets with single shear strength of 180 MPa. The allowable double shear force in a web-flange junction rivet: æ d2ö æ 3, 22 ö Fadm ,cis = 2* τ R * ç π ÷ = 2 *180 * ç π ÷ = 2895 N 4 ø è 4ø è

The allowable bearing force of the web (1,3 mm thick 2024-T3 element, σmat = 905 MPa) is: Fadm ,mat = σmat * e * d = 905 *1, 3 * 3, 2 = 3765 N The allowable double shear force is therefore more critical; the Reserve Factor is then: Fadm ,cis 2895 = = 1, 24 R.F = 2324 F CHECKING RIVETING OF REINFORCEMENT PLATE. Two rows of rivets with a rivet pitch of 38 mm. The load on two rivets is:

é T F = f *p = ê ë IA .N

ù

15 000

ò ydSúû *p = 2 587 790 * 5 315*38 = 1170 N

i.e. 585 N per rivet

Using the simplified formula, the load on the two rivets is:

é T Sp ù 15 000 70.4 * * 38 = 1198 N i.e. 599 N per rivet ú*p = 179.6 186.4 ë H Ss s û

F = f *p = ê

The rivets are stressed in single shear: æ d2 ö æ 3, 22 ö Fadm ,cis = τ R * ç π ÷ = 180* ç π ÷ = 1447 N 4 ø è 4ø è

The Reserve Factor is: R. F =


1447 585

= 2.47

Junctions

page V1-5•5–5




Junctions

Issue 1


V1-5 6 EXAMPLE Let us take an assembled beam with an I-section, bearing simply at 2 points and loaded by three forces concentrated as shown and which comprise the design loads for the beam. 200 pocket 28

630 Uprights made of sections: Dimensions :13,5*13,5*1

Span 4 eame = 1,5

11 000 N

pocket 22 pocket 21

630 90

Span 3 eame = 1,8

pocket 15 12 250 N

3,2 mm diameter rivets in flange, pitch 25 mm

pocket 14

630

Span 2 eame = 1,8

pocket 8 11 000 N

pocket 7 pocket 6 630

pocket 5 pocket 4

Span 1 eame = 1,5

pocket 3 pocket 2 pocket 1 Issue 1  AEROSPATIALE - 1999

Example

page V1-5•6–1

SSM V1-5 • STABLE WEB BEAMS 10,5 19

19

19

2,4

0,6 (sheet) 5,6 19 100 8

neutral fibre Sss = 85 mm2 Iss = 2856 mm4 upper flange

neutral fibre central line

200 web height 196,8

1,5

3,2 (rivet hole) 19

0,5 (skin)

11

Ssi = 58 mm2 Isi = 2020 mm4 neutral fibre 5.3 19

100

1.6

lower flange The characteristics of the materials: 2024 PLT 351 Skin, Sheet and Web; 7075 T73510 flanges Skin, Sheet and Web: σ R = 440 MPa ;σ 0,2 = 270 MPa ; cE= σ R = 495 MPa ;σ 0,2 = 420 MPa ; cE=

Flanges:

S mm2 170 22,8 295,2 354,24 116 50 - 13,44 - 15,04 - 16 625,5 683,6

ELEMENTS

upper flange upper skin web (th=1,5 mm) web (th=1,8 mm) lower flange lower skin flange-skin rivet holes flange-web rivet holes flange-web rivet holes TOTAL (1,5 mm web) TOTAL (1,8 mm web)

Y mm 94,4 100,3 0 0 - 94,7 - 100,25 - 99,45 - 89 - 89

Sy mm3 16048 2287 0 0 - 10985 - 5013 1337 1339 1424 5012 5098

70300 MPa ; n = 7.05 73800 MPa ; n = 13.83

Sy2 mm4 1514931 229370 0 0 1040298 502503 - 132926 - 119132 - 126736

I0 mm4 5712 negligible 952764 1143317 4040 negligible negligible negligible negligible

I0 + Sy2 mm4 1520643 229370 952764 1143317 1044338 502503 - 132926 - 119132 - 126736 3997561 4180509

Section with 1,5 mm web y=

Σ sy ΣS

=

5012 625,5

= 8 mm 2

IA.N. = (Σ I0 + Sy2) - (ΣS) y = 3997561 - 625,5 * 82 = 3957529 mm4 Section with 1,8 mm web: (with same position of neutral fibre) 2

IA.N. = (Σ I0 + Sy2) - (ΣS) y = 4180509 - 683,6 * 82 = 4136758 mm4


Example

Issue 1

SSM V1-5 • STABLE WEB BEAMS CHECKING BEAM BENDING STRENGTH 11000

17125

630

11000

12250

630

17125

630

630

16000000 14647500 14000000 12000000 10000000 8000000 6000000 4000000 2000000 x

0 630

1260

1890

2520

With the beam simply supported at both ends, with symmetrical loading and with a constant cross section along complete length. The maximum bending moment is found in the centre of the beam as shown by the equilibrium of the part and the bending moment diagram above. M max = 17125*1260 −11000 *630 = 14 647 500 mm.N The beam is riveted to a skin panel as beam section shows on previous page. We will associate a certain efficient sheet width to the upper flange which under the loading considered is in compression. This will depend on the stress in the flange of the beam. On tensioned flange side, the complete skin is load-carrying which corresponds to the half-distance to the first skin stiffener on each side (distance between 2 stiffeners: 100 mm). Details of the cross section in the centre of the beam are shown on the previous page with three rivet holes on the tensioned flange. The table on the same page gives the geometrical characteristics of the section in relation to the neutral fibre. The bending stress at the centre of gravity of the horizontal portion of the compressed flange is: σf =

Mf * v I

=

14 647 500 * 90.8 4 136 758

= 321.5 MPa

The critical stress in the 1 mm thick and 78 mm wide sheet (between fastener rows) is: kcπ2Ep æ e ö 4∗ π 2 ∗70300 æ 0,6 ö =η = η ç ÷ ç ÷ = 15.4 MPa 7 12(1 − ν 2 ) è b ø 12(1 − 0.332 ) è 78 ø 2

σ cp

2

Load-carrying width on a leg of the compressed flange: L t = L t0

σ cp σf

= 39

15.4 321.5

= 8,5 mm

The total width of each of the leg is 8,5 + 10,5 = 19 mm which is exactly the value taken for the calculation of the geometrical characteristics of the section of the beam.


Example

page V1-5•6–3

SSM V1-5 • STABLE WEB BEAMS CALCULATING ALLOWABLE COMPRESSION STRESS IN UPPER FLANGE The upper flange consists of two right angle extrusions with two equal legs attached both to the skin and to the web. The allowable stress of such a beam is the critical crippling stress which is equal to the critical buckling stress of one of the two equal legs of the section: å S i σ cri As per §V1-3.7: σ crip = = σ cr of one of the two legs. å Si σ cr = Kc

2

ηE c π 2

η * 73800 * π 2 æ eö ÷ = 0.43 2 ç 12(1 − ν e ) è b ø 12(1 − 0.332 )

2

æ 2.4 ö ç ÷ = η * 532.5 MPa è 1.8 ø

After plasticity correction σ cr = 417 MPa The Reserve Factor: R.F =

417 321.5

= 1.3

CHECKING LOWER FLANGE IN TENSION The bending stress at the centre of gravity of the skin is: M * v 14 647 500 * 108.25 = = 383.3 MPa σf = f I 4 136 758 440 The Reserve Factor (Skin): R.F = = 1.15 383.5 495 = 1.29 The Reserve Factor (Lower flange): R.F = 383.5 The Reserve Factor will be improved if we use the method based on the stress-strain diagram as explained in § 2.1.1 and which consists in calculating the ultimate bending moment. CHECKING BUCKLING STRESS IN WEB The maximum shear force is found at the bearing points and is equal to: 17125 N. The thickness of the web at the bearing points is 1,5 mm. The height of the web is the distance between the rows of fasteners connecting the web to the flanges: 178 mm. The spacing between the uprights is 90 mm. The maximum shear stress in the web can be calculated: T 17125 = = 64.1 MPa . by the simplified equation: τ = he 178 * 1,5 T 17125 by the exact equation: τ = ò ydS = 170 * 86.4 + 22.8 * 92.3 + 1,5 * 90.4 * 45.2 = 66.1 MPa . Ie 3957529 * 1,5 The bending moment is negligible near to the bearing points; therefore, we can consider the panels as being loaded with pure shear with the edges simply supported at the uprights and the fastener rows. 2 b æ 90 ö a = h = 178 mm. b = 90 mm. = 0, 5 Þ k s = 3, 8 * ç ÷ + 5, 35 = 6, 3 see §V1-2.2.3.1.2 a è 178 ø

[

2

τ cr,0

]

2

ηk sπ 2 E æ e ö 6,3 * π 2 * 70300 æ 1,5ö = =η ç ÷ ç ÷ = η * 113.5 MPa = 107 MPa (h = 0.943). 12 * (1 − 0,332 ) è 90 ø 12 1 − ν 2e è b ø

(

)

The Reserve Factor is: R.F. =

107 66,1

= 1.62

This value is fairly correct as the real boundary condition for the web is certainly more stabilising than the one taken, that is simple supported.


Example

Issue 1

SSM V1-5 • STABLE WEB BEAMS CHECKING END SPANS WITH A WEB THICKNESS OF 1,5 MM. The most critical webs are those of pockets 7 and 22. They are submitted to a shear load of 17125 N superimposed on a bending moment in the centre of the pocket equivalent to: 17125 * 585 = 10 018 125 mm.N. As the web is clamped between the sections of the upper flange and those of the lower flange, the buckling of the web will occur near to the lower edge of the upper flange sections, that is, at a distance of 73 mm from the neutral fibre. M * v 10 018 125 * 73 The bending stress is: σ f = f = = 184.8 MPa I 3 957 529 ηk f π 2E æ e ö 2 The critical buckling stress in bending: σ cf,0 = 2 ç ÷ . 12(1 − νe ) è b ø The height of the pocket h=b = 162 mm. The length of the pocket = spacing between uprights = a = 90 mm. a 90 = = 0, 55 Þ k f = 41, 8 (see §V1-2.2.3.1.3) pocket with two sides clamped and two others bearing. b 162 σ cf,0 =

η * 41,8 * π 2 * 70300 æ 1,5 ö

(

12 * 1 − 0,332

2

ç ÷ = η * 232.5 MPa è 162 ø

)

The plasticity correction factor for a plate with two clamped edges and two others hinged is given in § V1-2.2.2 : η = η4 = 0.856 henceσ cf,0 = 199 MPa . The shear stress at pocket 7 or 22 is the same as the one calculated at the bearing points, as the shear force is the same. Thus τ = 66.1 MPa ;τ cr,0 = 107 MPa. The interaction equation for the combined bending - shear loads is: (see § V1-2.2.4.1 ) σ 184.8 66.1 τ 2 2 R f + R s = 1 with R f = f = = 0,93 and R s = = = 0.62 199 σ cf,0 τ cr,0 107

R.F. =

1 2 Rf

+

2 Rs

=

1 0.93 + 0.62 2

2

= 0,9

CHECKING SPANS IN CENTRE OF BEAM WITH A 1.8 MM WEB. The bending moment at the centre of a pocket adjacent to the centre of the beam (pockets 14 and 15) is equal to: M f = 17125*1215− 11000 *585 = 14 371 875 mm.N. The shear force on the same pocket is equal to: T= 17125− 11000 = 6125 N M * v 14 371 875 * 73 The bending stress is: σ f = f = = 253.6 MPa I 4 136 758 The shear stress is: T 6125 τ = ò ydS = 170 * 86.4 + 22.8 * 92.3 + 1.8 * 4086 = 19.9 MPa Ie 4136758 * 1.8

[

The critical buckling stress in bending: σ cf,0 =

]

η * 41.8 * π 2 * 70300 æ 1.8 ö

(

12 * 1 − 0.332

)

2

ç ÷ = η * 334.8 MPa è 162 ø

The plasticity correction factor for a plate with two clamped edges and two others hinged is calculated as per § V1-2.2.2 : η = η4 = 0.708 hence σ cf,0 = 237 MPa . The critical shear stress: 2

τ cr,0

2

ηk s π 2 E æ e ö 6,3 * π 2 * 70300 æ 1.8 ö = =η ç ÷ ç ÷ = η * 163.5 MPa = 133.2 MPa (η = 0.815). 12 * (1 − 0.332 ) è 90 ø 12 1 − ν 2e è b ø

(

)


Example

page V1-5•6–5

SSM V1-5 • STABLE WEB BEAMS The interaction equation for the combined bending - shear loads is: (see § V1-2.2.4.1 ) R 2f + R 2s = 1 where R f = R.F. =

1 Rf + Rs 2

2

σf 253.6 = = 1,07 237 σ cf ,0 1

=

1.07 + 0.15 2

2

and

Rs =

τ τ cr,0

=

19.9 133.2

= 0.15

= 0,93

CHECKING RIGIDITY OF UPRIGHTS. The web is stiffened on a single side by the uprights: sections with legs of equal size (mm) 13,5*13,5*1. The formula of § 5 gives the moment of inertia required for an upright to prevent buckling. 4

4

2, 29 l æ Tcr h a ö 3 2, 29 *90 æ 17125*196, 8ö 3 Im = ç ç ÷ = ÷ = 204 mm 4 e è 33 E ø 1, 5 è 33*70300 ø The moment of inertia of the uprights is 460 mm4, therefore greater than the moment of inertia required. Such uprights are therefore satisfactory. The moment of inertia required will also be checked by using the curve on page V1-5.5.2: l 90 l The parameter on the X-axis is: = 0,5 = 0 ,5 = 0.21 0 ,5 2 h * Ks æ π2 * k ö æ π * 6, 3 ö s h * çç 178* çç ÷÷ ÷÷ 2 2 è (1 − ν e )* 12 ø è (1 − 0, 33 )*12 ø I 3 4 From the curve, we obtain m3 ≈ 1, 4 Þ Im = 1, 4 * 90 *1, 5 = 425 mm . le The rigidity of the uprights is therefore satisfactory. CHECKING RIVETS ATTACHING THE WEB TO THE FLANGES, END SPANS.

In the end spans

The shear load T = 17125 daN Web thickness e = 1,5 mm

The shear flow is:

T 17125 = = 85.6 N/mm H 200 T 17125 by exact formula: f = ò ydS = (85 * 2 * 86.4 + 19 * 2 * 92.3) = 78.7 N/mm I 3957529 All rivets are made of aluminium alloy 2117-T3, diameter 3,2 mm with a pitch of 25 mm and a shear strength of 180 MPa. by approximate formula: f =

æ π * d2 ö The allowable single shear force is then: Fadm cis = 180* ç ÷ = 1448 daN è 4 ø As the web-flange junction rivets are submitted to double shear: F adm cis = 2896 daN The maximum force in a rivet is: f * pitch = 85.6 * 25 = 2140 daN, which gives R.F = 1.35. The rivets attaching the skin to the upper flange must have a pitch so as to prevent all buckling of the skin between the rivets as we retained a certain efficiency of the skin in the beam moment of inertia calculation. However, the rivets attaching the skin to the lower flange need not be checked for buckling between rivets, as the skin is tensioned, but simply for shear loading. The uprights can be checked at the external load application points and at the beam bearing points as they transfer the concentrated loads to the web of the beam.


Example

Issue 1


V1-5 7 STABLE WEB BEAMS WITH ROUND LIGHTENING HOLES V1-5 7.1 GENERAL The type of web analysed in the previous paragraphs includes many elements (uprights) to obtain the required lightness. To reduce manufacturing costs, the number of parts can be reduced. A trade-off must be found between the manufacturing costs and excess weight. This trade-off often depends on the type of aircraft under calculation. Thus, reducing the number of elements enables extensive savings in manufacturing costs and can thus be more economical. To reduce the number of uprights, we frequently use a web drilled with round flanged lightening holes with varied spacing. However, there is a general limitation to this as we must place an upright at locations where high loads are applied to the beam. Webs with holes provide many accesses for hydraulic and electrical conduits. Load

b A

A Flange

Upright under concentrated load d

C

B

C’

e D

H

g Section A-A

V1-5 7.2 STANDARDISATION OF FLANGED HOLES Design standards at Aérospatiale define the geometry of flanged holes according to the diameter of the hole (Standard ASN 451.01). Thus, we will give below the values of: d, D, H (in mm). e = 0,6 - 0,8 - 1 - 1,2 - 1,4 d D H 15 20 3 25 30 3 30 36 3 40 45,5 3 50 55,5 3 60 65,8 3,5 70 75,8 3,5 80 85,8 3,5 90 96,1 4 100 106,3 4 110 120 5 120 133,6 6 130 143,8 6 140 153,6 6 150 163,7 6 160 173,6 6


e = 1,6 - 1,8 - 2 - 2,5 d D 15 22,7 25 32,7 30 37,7 40 48,3 50 58,3 60 68,7 70 78,7 80 88,7 90 98,9 100 109 110 123 120 136,8 130 146,9 140 157 150 166,9 160 177

Stable Web Beams With Round Lightening Holes

H 4 4 4 4 4 4,5 4,5 4,5 4,5 5 6 7 7 7 7 7

page V1-5•7–1

SSM V1-5 • STABLE WEB BEAMS V1-5 7.3 ALLOWABLE SHEAR FLOW The rule-of-thumb formula giving the allowable shear flow (in relation to failure) for webs with these types of holes is: é d ù C′ 2 fadm = k × e × êτ h 1 − (d / b) + τ c ú× bû g ë

(

)

k = 0.85 − 0.0006 × (b / e)

τ h or τ c = shear stress producing the collapse of a long plate with width h or C and thickness e (obtained for instance by using the curves on the following page). g = distance between the centrelines of two adjacent holes. b = height of the web between the lower and upper fastener rows (web - flange fastener). B=

D−d 2

and C = g - d

C’ = C - 2B = C + d - D = g - D

In general, we find, with the rule-of-thumb formula, that the largest holes (diameter around 0.8 h) with a fairly large spacing g give lighter but less rigid webs. In addition to the calculation above (collapse), it must also be checked at the holes that the web under ultimate load f × b does not give net shear stresses above the breaking shear τ R . Calculation graphs can be plotted from the rule-of-thumb formula (see example on page V1-5.7-4).



Issue 1


Zh, Zc en MPa

Zh, Zc en MPa 140

280

126

252

112

238

98

224

84

210

70 (e) mm 0,4

196 182

42

0,5 0,6 0,8 1,0 1,3 1,6

168 154

140

56

28 14

NACA A.R.R. DEC., 1942 0 0

20

40

60

80

100

120

140

160

180

200

220

240

260

280

page V1-5•7–3

CONTRAINTES DE CISAILLEMENT DE RUINE, Zh ou Zc1 POUR LES AMES EN ALLIAGE D'ALUMINIUM 2024

300

320



266


Flux de cisaillement admissible (N/mm) 157.5 h = 100

d/h = .50

h = 100

140


g = 3.0 d

122.5 g = 2.25 d

105

h = 100

h = 100

87.5 175

175

175

250

250

175

250

250

h = 100

70

h = 100 g = 1.5 d

52.5

175 175 250

35

250

17.5 NACA A.R.R. DEC., 1942 0 Issue 1

0,5

1,0

1,5

0,5

1,0

1,5

0,5

1,0

1,5

épaisseur de l'âme (mm) FLUX DE CISAILLEMENT ADMISSIBLE (EFFONDREMENT) POUR DES AMES EN 2024 AVEC TROUS D'ALLEGEMENT CIRCULAIRES PRESENTANT DES BORDS TOMBES A 45°


d/h = .80

SSM V1-5 • STABLE WEB BEAMS V1-5 7.4 REMARKS 3 options are possible: 1. No uprights:

Stable web (plates of infinite length). E.g.: Extrusions.

2. Uprights:

Stable and thin iso-loaded web (plates of finite length). The uprights must be correctly sized.

3. Flanged holes:

Lightening - stabilisation trade-off. For safety reasons, we place the uprights at the concentrated loads and in areas with high distributed loads to avoid crushing.

V1-5 7.5 EXAMPLE Let us consider the beam shown below (same geometry and design loads as the one of example §5)

h = 178

11000

12250

11000

Span "A"

Span "B"

Span "B"

Span "A"

630

630

630

630

17125

17125

Let us dimension the webs drilled with round lightening holes. Using the simplified formula, we can find the shear flux in spans A and B: fA =

T H

=

17125 200

= 85.6 N / mm and f B =

T H

=

17125 − 11000 200

= 30.6 N / mm

Thus span "B" will be lighter than span "A". The introduction of an 11000 N load means that at this location a load upright must make the junction between spans "A" and "B".

As large-diameter holes (and with lower spacings) are more efficient for a gain in weight, let us consider two holes of diameter d ≈ 0.8 h in each span (of 630 mm) spaced 315 mm apart. We thus obtain the following values: h = 178 mm Þ d = 140 mm Þ D = 153, 6 mm and H = 6 mm (table page V1 - 5.7.1)

g = 315 mm Þ C = g - d = 315 -140 = 175 mm and C'= g - D = 315 -153, 6 = 161, 4 mm



page V1-5•7–5

SSM V1-5 • STABLE WEB BEAMS Span "A" Assuming a web thickness of 2 mm. h 178 h = = 89 Þ k = 0, 85- 0, 0006 * ( ) = 0, 85 - 0, 0006 * 89 = 0, 8 e e 2 C 175 h 178 From the curves on page V1-5.7.3 for = = 89 and = = 87, 5 we obtain: e 2 e 2 τ h = 92.5 MPa et τ c = 94.5 MPa The allowable shear flow is then: é æ æ d ö2 é æ æ 140 ö 2 ö 140 ù 161, 4 d ö ù C' * = 0, 8 * 2 * 92, 5 * ÷ ÷ + 94, 5 * = 97, 6 N/mm fadm = k * e * êτ h ç1 − ç ÷ + τ c ú ê ú* ÷ ç1 − ç 178 û 315 h øû g è è 178 ø ø ë è èhø ë The Reserve Factor for collapse is: R.F =

f adm fA

=

97.6 85.6

= 1.14

Let us check the pure shear stresses in a section at a hole: T 17125 τ net = = = 225 MPa * e h − d 178 ( ) ( −140) * 2 The breaking shear for 2 mm thick material 2024 PL T351 is 270 MPa. The pure shear Reserve Factor is: R.F =

τR τ net

=

270 225

= 1.2

Span "B" Assuming a web thickness of 1,3 mm. h 178 = = 137 Þ k = 0, 85 - 0, 0006 * (h/e) = 0, 85 - 0, 0006 * 137 = 0, 77 e 1, 3 h 178 C 175 From the curves on page V1-5.7.3 for = = 137 and = = 135 we obtain: e 1, 3 e 1, 3 τ h = 59.5 MPa and τ c = 60.5 MPa The allowable shear flow is then:

fadm

ù é æ æ 140 ö 2 ö é æ æ dö 2 d ö ù C' ÷ + 60.5 * 140 ú * 161.4 = 39.2 MPa ç1 − ç ê ç ÷ ú = k * e * êτ h ç 1 − ç ÷ + τ c * = 0.77 * 1,3 * 59.5 * ÷ ç è 178 ø ÷ h ÷ø ú g 178 ú 315 ê êë è è h ø ø è û û ë

The Reserve Factor for collapse is: R.F =

f adm fB

=

39.2 30.6

= 1.28

Let us check the pure shear stresses for a section at a hole: T 6125 τ net = = = 124 MPa (h − d) *e (178 −140) *1,3 The pure shear Reserve Factor is: R.F =

τR τ net

=

270 124

= 2.18

Remarks a) The stressman can use the procedure above for smaller holes and lower thicknesses to find an arrangement making the web lighter. b) We can see that the examples above, for calculating the allowable flows, are confirmed by the curves on page V1-5.7.4. (the curve for the 2 mm thick web requires an extrapolation). c) The beam equipped with vertical uprights studied in example of § 5 is much heavier than the beam studied above.



Issue 1


YES

X

NO


1 4

5 4

2 1 2

Volume 1

5


Purpose

Scope


Contents

Document Manager



Not applicable.

V1 - 1 V1 - 2 V1 - 3 V1 - 4 V1 - 5 V1 - 6 V1 - 7 V1 - 8 V1 - 9



Validation

Name: J. HUET



3page 1

Title - Annex

Reference documents


Abbreviations




Definitions


Highlights

Issue

Date

Pages modified

A

02/98

V1 - 1 à V1 - 3 V1 - 7 à V1 - 9

B

05/99

V1 - 7


V1 - 4

New chapter.

V1 - 1


C

11/99




New chapter.

MTS 004 Iss. C

3Ann. page



List of approval

Dept. code

Function

BTE/CC/CM

Name / First name


Key words

Calcul

Bibliography

Néant

Signature

CAZET G.

Distribution list

Dept. code

Function


BQP/TE

Archives Diderot

SIBADE Alain

BQP/TE


SIBADE Alain


BOUTET Fernand

BTE/SM/MG



MTS 004 Iss. C

page IG1

SSM V1-7 • LUGS CONTENTS V1-7 V1-7•1 V1-7•2 V1-7•3 V1-7•4 V1-7•5 V1-7•6 V1-7•7 V1-7•8

LUGS GENERAL DATA PREPARATION ALLOWABLE AXIAL FORCE ALLOWABLE TRANSVERSE FORCE STRENGTH UNDER AN OBLIQUE FORCE ALLOWABLE BEARING IN BUSH PIN STRESSING NUMERICAL EXAMPLE

issue 1 1 1 1 1 0 1 1 1

date 4/1999 4/1999 4/1999 4/1999 4/1999 1/1998 4/1999 4/1999 4/1999

revision

V1-7 1 GENERAL

1-1

V1-7 1.1 INTRODUCTION - GENERAL PRESENTATION

1-1

V1-7 1.2 PRINCIPAL DIRECTIONS

1-2

V1-7 V1-7 V1-7 V1-7

1-3 1-3 1-4 1-5

1.3 FAILURE MODES 1.3.1 Failure under axial load 1.3.2 Failure under a transverse load 1.3.3 General information on pin stressing

V1-7 2 DATA PREPARATION

2-1

V1-7 2.1 MATERIAL PROPERTIES

2-1

V1-7 2.2 GEOMETRICAL PROPERTIES

2-1

V1-7 3 ALLOWABLE AXIAL LOAD V1-7 V1-7 V1-7 V1-7 V1-7 V1-7

3-1

3.1 ALLOWABLE TENSION LOAD 3.1.1 Determination of the elastic over-stress factor 3.1.2 Neuber rule 3.1.3 Application of the Neuber rule to the allowable tension load calculation 3.1.4 Calculation of εs 3.2 ALLOWABLE SHEAR/BEARING LOAD

3-1 3-1 3-1 3-2 3-3 3-5

V1-7 3.3 SPECIAL CASE CONCERNING EYE ENDS

3-7

V1-7 3.4 BLOCK DIAGRAM V1-7 3.4.1 Solid lugs V1-7 3.4.2 Eye ends

3-11 3-11 3-12

V1-7 4 ALLOWABLE TRANSVERSE LOAD

4-1

V1-7 5 STRENGTH UNDER AN OBLIQUE LOAD

5-1

Issue 1 © Copyright AEROSPATIALE 1998

General

page V1-7•i

SSM V1-7 • LUGS V1-7 6 ALLOWABLE BEARING

6-1

V1-7 7 PIN STRESSING

7-1

V1-7 7.1 PIN SHEAR

7-1

V1-7 V1-7 V1-7 V1-7

7-1 7-1 7-3 7-12

7.2 PIN BENDING 7.2.1 Calculation of the lever arm "b" 7.2.2 Calculation of the bending modulus 7.2.3 Allowable bending load

V1-7 8 INTERACTIVE CALCULATION

8-1

V1-7 8.1 PRINCIPLE

8-1

V1-7 8.2 EFFECT ON ALLOWABLE AXIAL LOAD

8-2

V1-7 8.3 EFFECT ON ALLOWABLE TRANSVERSE LOAD

8-3

page V1-7•ii © Copyright AEROSPATIALE 1998

General

Issue 1

SSM V1-7 • LUGS SYMBOLS USED Abbreviation CG UL LL RF

Meaning centre of gravity ultimate load limit load reserve factor

Notation a A Aav Abr Abrb As At A1 A2

Unit (mm) (mm²) (mm²) (mm²) (mm²) (mm²) (mm²) (mm²) (mm²)

A3

(mm²)

A4

(mm²)

b C d di D e E Ea F Fa Fbru Fbry Ftr Ftux Ftyx g I I2/I3 k K Kbr Kbrv Kte Ktru Ktrv Kt Kε Kσ L

(mm) (mm) (mm) (mm) (mm) % (daN/mm²) (daN/mm²) (daN) (daN) (daN/mm²) (daN/mm²) (daN) (daN/mm²) (daN/mm²) (mm) (mm4) (mm4)

Mf

(daN.mm)

(mm)

Description distance from centre of hole to edge of lug in axial direction cross section area of eye end ring weighted surface area for calculation of lug subjected to transverse force projected bearing surface area minimum bearing surface area at the pin/bush/bore interface sheared surface area critical tensile cross section area cross section area of lug subjected to transverse force in a plane 45° to pin hole axis 1) (axial strength of eye ends) cross section area of ring at point ② 2) (transverse strength, all lugs) cross section area of lug subjected to transverse force at pin hole 1) (axial strength of eye ends) cross section area of ring at point ③ 2) (transverse strength, all lugs) minimum radial cross section area cross section area of lug, taken opposite force in transverse direction in a plane 45° to pin hole lever arm curved abscissa (eye ends) pin diameter pin inside diameter hole diameter minimum elongation at break lug modulus of elasticity pin modulus of elasticity force applied at ultimate load projection of F (U.L.) or F/1.5 (L.L.) in axial direction bearing stress ultimate strength (e/D = 2) conventional bearing yield strength (e/D = 2) projection of F (U.L.) or F/1.5 (L.L.) in transverse direction allowable tensile limit at ultimate load allowable tensile limit at ultimate load clearance between male and female parts moment of inertia of a section in relation to its own centre of gravity moment "I" at points② and ③ for eye ends in tension section geometrical factor dimensionless factor used for eye end calculations shear/bearing factor at ultimate load shear/bearing factor at limit load elastic tensile overstress factor transverse factor at ultimate load transverse factor at limit load AEROSPATIALE FATIGUE MANUAL overstress factor real strain concentration factor real overstress factor distance from centre of hole to edge of lug in a direction parallel to axis of symmetry of external profile pin bending moment


General

page V1-7•iii

SSM V1-7 • LUGS M3 N2/N3 Padm Pbru Pbrv P' brv Pfu Pfv Psu Ptru Ptrv Ptu Ptv Ptu2

(daN.mm) (daN) (daN) (daN) (daN) (daN) (daN) (daN) (daN) (daN) (daN) (daN) (daN) (daN)

Ptv2

(daN)

Ptu31

(daN)

Ptv3

(daN)

Ptu32

(daN)

Pu Pv r

(daN) (daN) (mm)

R R3 Rtr t t' 1/t' 2 T3 v v3 W

(mm)

β, ϕ, ψ γ εMAX εeMAX εpMAX εmax εnom εs ε0.2 θ σapp σb σbrg σbru σbrv σe max σmax σMAX or σM σnom

(mm) (mm) (daN) (mm) (mm) (mm) (°)

(°) (daN/mm²) (daN/mm²) (daN/mm²) (daN/mm²) (daN/mm²) (daN/mm²) (daN/mm²) (daN/mm²) (daN/mm²)

bending moment at point ③ for eye ends in tension normal force at points ② and ③ for eye ends in tension allowable oblique force on lug allowable shear/bearing force at ultimate load allowable shear/bearing force at limit load allowable bush bearing force allowable bending force at ultimate load allowable bending force at limit load allowable pin shear force at ultimate load allowable transverse force at ultimate load allowable transverse force at limit load allowable tensile force at ultimate load allowable tensile force at limit load allowable tensile force at ultimate load conditioned by static strength at point ② (eye ends) allowable tensile force at limit load conditioned by static strength at point ② (eye ends) allowable tensile force at ultimate load conditioned by static strength under normal stress at point ③ (eye ends) allowable tensile force at limit load conditioned by static strength at point ③ (eye ends) allowable tensile force at ultimate load conditioned by static strength under tangential stress at point ③ (eye ends) allowable axial force at ultimate load allowable axial force at limit load 1) neutral fibre radius of an eye end ring 2) (peaking) dimensionless quantity used for calculatingγ lug nose radius ratio of Fa over Pu (U.L.) or Fa over Pv (L.L.) ratio of Ftr over Ptru (U.L.) or Ftr over Ptry (L.L.) thicknesses of lug parts reduced active thicknesses of lug shear force at point ③ for eye ends in tension distance between CG and extreme fibre of a section dimension v at point ③ for eye ends in tension lug width angles used in calculating eye ends subjected to axial force peaking coefficient maximum strain corresponding to σMAX maximum allowable elastic strain maximum allowable plastic strain maximum strain corresponding to σmax strain corresponding to σnom strain corresponding to start of striction strain corresponding to σ0,2 angle between axial direction and direction of application of F bending stress in pin at extreme fibre bending modulus bearing stress in lug allowable bearing stress 0.2% bearing yield strength maximum elastic tensile stress maximum real tensile stress in lug maximum bending stress in pin (at neutral fibre) nominal tensile elastic stress in the critical section of lug

page V1-7•iv © Copyright AEROSPATIALE 1998

General

Issue 1

SSM V1-7 • LUGS σR σ0 σ0,2 σc0,2

σ2/σ3 τ τadm τ3 Index 1 2

(daN/mm²) (daN/mm²) (daN/mm²) (daN/mm²) (daN/mm²) (daN/mm²) (daN/mm²) (daN/mm²)

ultimate strength Cozzone's fictitious stress at neutral fibre in pin conventional tensile yield strength conventional compression yield strength normal stress at points ② and ③ for eye ends in tension shear stress allowable pure shear stress tangential stress at point ③ for eye ends in tension Meaning characteristic index of female part of assembly characteristic index of male part of assembly (except eye ends)


General

page V1-7•v

SSM V1-7 • LUGS


page V1-7•vi © Copyright AEROSPATIALE 1998

General

Issue 1

SSM V1-7 • LUGS REFERENCES

1 - AIRFRAME STRUCTURAL DESIGN, Nyu, Michael C. Y. 2 - MANUEL FATIGUE AEROSPATIALE MTS 005 3 - LUGS AND SHEAR PINS, Melcon, M. A. & Hoblit, F. M. in Product Engineering, June 1953 4 - FORMULAS FOR STRESS, STRAIN AND STRUCTURAL MATRICES, Pilkey, Walter D. 5 - ANALYSIS & DESIGN OF FLIGHT VEHICLE STRUCTURES, Brühn, E. F. 6 - STRESS MEMO No. 1e, Lockheed-California Company. 7 - SUPERSEDES SECT. 7.7, Canadair.


General

page V1-7•vii

SSM V1-7 • LUGS


page V1-7•viii © Copyright AEROSPATIALE 1998

General

Issue 1

SSM V1-7 • LUGS V1-7 1

GENERAL

V1-7 1.1 INTRODUCTION - GENERAL PRESENTATION Traditionally, lug junctions are used in a certain number of cases: - attachment of engine pylons under the wing - rods inside wing boxes - inter-rail floor beams - floating floor junctions with adjacent zones - door fittings - galley upper attachments - etc. Generally, a lug-type junction comprises: - a male part - a female part - a pin More rarely, a junction consisting of two male parts is found. The bores in the male and female parts may be fitted with bushes to avoid direct contact with the pin.

Female part of the lug junction

Pin

Male part of the lug junction

Figure V1-7 .1.1-1: Example of a lug assembly As these junctions often transfer single forces, special care must be paid to sizing, especially by taking into account the fitting factor, imposed by certification texts. Most often, the following is recommended: - take either a fitting factor of 1.15 which increases the force applied: - or a margin of at least 20%.


General

page V1-7•1-1

SSM V1-7 • LUGS Note that these special precautions apply to all items for which mechanical strength is not guaranteed by tests. This may concern the male and female parts as well as the pin. A calculation at ultimate load and a calculation at limit load are made. The limit load calculation is especially important when the removability of the junction must be conserved for maintenance operations.

V1-7 1.2 PRINCIPAL DIRECTIONS Before calculating this type of junction, it is essential to define two main force directions. In the simple case where the hole is centred on the bisector of the external profile of the lug (refer to figure V17 .1.2-1): - the axial direction is defined as being this bisector, - the transverse direction is defined as being perpendicular to the axial direction.

Bisector Axial direction

Figure V1-7 .1.2-1: Lug with centred hole

In certain cases, the hole is not located on the bisector. Depending on the distance between the centre of the bore and the edge of the part, the axial direction may be plotted either in a direction parallel to the bisector or in the minimum material direction (refer to Figure V1-7 .1.2.2).

R

R

Bisector

Bisector Axial direction L

L

Axial direction

Figure V1-7 .1.2-2: Lugs with asymmetrical holes

page V1-7•1-2 © Copyright AEROSPATIALE 1998

General

Issue 1

SSM V1-7 • LUGS Example for determining axial direction: adapter beam end lug

Axial direction

Tangent to the outer profile Outer profile Bisector Figure V1-7 .1.2-3: Example for determining axial direction In the example of Figure V1-7.1.2-3, the hole is located on the bisector. Therefore, the axial direction coincides with this bisector. The transverse direction is perpendicular to the axial direction, orientated in the orthogonal projection direction of the applied force, downwards or upwards.

V1-7 1.3 FAILURE MODES V1-7 1.3.1 Failure under axial force The following may occur under an axial force: - tensile failure: the concentration of stresses at the edge of the hole cause failure in a section approximately perpendicular to the axial direction and located on either side of the hole (Figure V17.1.3.1-1).

F

Figure V1-7 .1.3.1-1: Tensile failure under axial force - failure by a combination of bearing and shear: the bearing and shear stresses combine to cause failure in planes located at ± 40° to the axial direction (Figure V1-7.1.3.1.2).

40° 40°

F

Figure V1-7 .1.3.1-2: Bearing/shear failure Issue 1 © Copyright AEROSPATIALE 1998

General

page V1-7•1-3

SSM V1-7 • LUGS In the specific case of eye ends, failure under axial load occurs generally at the junction between the lug ring and the remainder of the part (refer to Figure V1-7.1.3.1-3). Essentially, this is a failure caused by annular tensile stresses.

F

Figure V1-7 .1.3.1-3: Axial failure of an eye end

V1-7 1.3.2 Failure under transverse force The failure mode is more complex under a transverse force. Failure is the combination of several modes, including: - material shear between the hole and the outer edge of the lug, where the force is applied, - local bending of the same segment of material, - circumferential tension in the annular ring of material between the hole and the outer edge of the lug. Furthermore, a weakness in one of these modes is compensated for by all the others. Locally the load paths may vary. The generic term "transverse failure" is used to designate this combination.

F F

Shear at load

Bending + tensile failure due to annular stress

Figure V1-7 .1.3.2-1: Examples of transverse failures

Generally, the force is not purely axial or purely transverse. Therefore, it is called an "oblique force" comprising both an axial component and a transverse component. The calculation procedures given here make it possible initially to determine the allowable axial and transverse forces, with the possibility of combining them to calculate a margin for an oblique force. Then, we calculate the strength of the lug pin.


General

Issue 1

SSM V1-7 • LUGS V1-7 1.3.3 General information on pin stressing The pin is substantiated in shear and bending. The calculation of the real lever arm for pin bending takes the "peaking" phenomenon into account: bending strain causes a concentration of contact pressures close to the outer faces of the male part and the inner faces of the female part. F/2

F/2

F/2

F/2

Female part

Inactive sections on thicknesses Male part F

F

CONCENTRATION OF CONTACT PRESSURES (PEAKING)

UNIFORM DISTRIBUTION OF CONTACT PRESSURES

Fig. V1-7 .1.3.3-1: Distribution of contact pressures in the lug The bending calculation also takes into account the plasticization of the extreme fibres of the pin using the Cozzone method.


General

page V1-7•1-5

SSM V1-7 • LUGS



General

Issue 1

SSM V1-7 • LUGS

V1-7 2

DATA PREPARATION

V1-7 2.1 MATERIAL CHARACTERISTICS The following information on the male and female parts of the lug junction is required: - the ultimate tensile strength: σR - the ultimate bearing strength (e/D = 2): Fbru - the conventional bearing yield strength (e/D = 2): Fbry - the conventional tensile yield strength: σ0.2 - the conventional compression yield strength: σc0.2 - the modulus of elasticity in tension: E - minimum elongation at break c(%) As far as possible, the following shall be obtained: - the factor "n" of the Ramberg & Osgood equation (tensile curve) For an eye end: - the allowable pure shear stress: - the ultimate tensile strength: - the conventional tensile yield strength:

τR σR σ0.2

If the assembly includes a bush, record: - the conventional compression yield stress:

σc0.2

The following information is required for thepin: - the ultimate strength: σR - the conventional yield stress: σ0.2 - the conventional compression yield strength: σc0.2 - the allowable pure shear stress: τadm - the modulus of elasticity in tension: Ea - the factor "n" of the Ramberg & Osgood equation (tensile curve)

V1-7 2.2 GEOMETRICAL CHARACTERISTICS In addition to the thickness, the dimensions to be recorded for the lug calculations are shown on Figures V1-7 2.2-1 to -4. R

W Axial direction

L=a

Fig. V1-7 .2.2-1: Lug with parallel edges


Data preparation

page V1-7•2-1

SSM V1-7 • LUGS

R

W/2

Axial direction

β

W/2 L=a Fig. V1-7 .2.2-2: Lug with converging edges

R

Axial direction

W/2 L=a Fig. V1-7 .2.2-3: Lug with hole off-centred in relation to the outer profile and large edge distance (L>R) R a

W/2

L

Axial direction

Fig. V1-7 .2.2-4: Lug with hole off-centred in relation to the outer profile and small edge distance (L

Data preparation

Issue 1

SSM V1-7 • LUGS The following indices will be associated with all specific quantities either for the male part or for the female part: - "1" for the male part, - "2" for the female part. For a "male-male" assembly, index 2 concerns the second male part. Two characteristic sections will be used for the lug calculations: - the projected bearing section: A br = D.t - the net section: At = (W – D)t = æç W − 1ö÷ .A br èD ø Another surface called Aav is used for calculations on lugs subjected to transverse forces; as it is complex to determine, all the necessary details are given in § V1-7 .4.

If the assembly includes a bush, also use the bearing surface area in the bush, called Abrb, defined as follows: Abrb = d.t For pin calculations, the following are required: - the diameter of the pin: - the maximum clearance between the male and female parts: For a hollow pin, the inside diameter will be called di.


Data preparation

d g

page V1-7•2-3

SSM V1-7 • LUGS



Data preparation

Issue 1


ALLOWABLE AXIAL FORCE

The allowable axial force at U.L. is called Pu; this is the minimum value between the allowable tensile force (Ptu) and the allowable shear/bearing force (Pbru): Pu = min{Ptu; Pbru} The allowable axial force at L.L., Py is determined globally from Pu. Ftyx Py = Ky. .Pu Ftux Ftyx = 0.593.Fbry F Ftux= bru 1.9 Pu Pu Si ≥ 1.05: Ky = 1.3625 - 0.25. A br .Ftux A br .Ftux

Si

Pu ≤ 1.05: Ky = 1.1 A br .Ftux

V1-7 3.1 BREAKING FORCE IN TENSION Axial forces create a concentration of the stresses at hole edge. This concentration can be sufficiently high to plasticize the metal. Therefore, to calculate the real amplified stress, the elastic overstress factor must be weighted with a plasticity correction. The procedure to be followed to calculate the allowable tensile force is given below.

V1-7 3.1.1 Determining elastic overstress factor The elastic overstress factor, called Kte, is used to calculate the overstress at hole edge on the basis of a purely elastic behaviour of the material. Thus: σe max = Kte . σnom where σnom is the nominal reference stress calculated considering net section At = (W-D).t. Refer to the fatigue manual (MTS 005 ch.III.3.8) for determining Kt: Kt = KtET.G The reference section used in the fatigue manual is the gross section, that is Abrut = W.t Thus: At Kte = .Kt A brut where Kt is the elastic overstress factor directly obtained from the fatigue manual graphs. Kte =

W −D .Kt W

V1-7 3.1.2 Neuber rule The Neuber rule is used to calculate a real overstress in the plastic range using the reference stress status, the elastic overstress factor and the stress - strain curve of the material. Normally, a given force is used to calculate the real stress and possibly calculate a margin in relation to the ultimate strength of the material. Let: - σnom be the uniform stress used as reference: σnom =

F At

- and εnom the corresponding strain.


Allowable axial load

page V1-7•3-1

SSM V1-7 • LUGS The equation of a hyperbola is defined assuming that the product of the real stress concentration factor Kσ and the real strain concentration factor Kε is equal to the square of the theoretical stress concentration factor (elastic) K te, let: σ where: Kσ = max Kσ . Kε = K2te σ nom ε max Kε = ε nom σ e max Kte = σ nom σe max is a fictious amplified stress as it is calculated on the basis of a purely linear elastic behaviour of the material.

Thus, we obtain:

σmax . εmax = K2te . σnom . εnom

Knowing Kte, σnom and εnom, the previous equality can be written:

σmax . εmax = Cte

C te . By plotting this hyperbola in the same plane ε as being the coordinates of the intersection point between the

This equality enables us to define an equation hyperbola:σ = as the material curve, we determine σmax and εmax two curves (refer to Figure V1-7 .3.1.2.-1). σ

σe max σmax

Neuber hyperbola

Material curve σnom

εnom

εmax

Fig. V1-7 .3.1.2-1: Graphic interpretation of the Neuber rule page V1-7•3-2 © Copyright AEROSPATIALE 1998


Issue 1

SSM V1-7 • LUGS V1-7 3.1.3 Applying Neuber rule to allowable tensile force calculation Generally, for lug calculations, we try to express the capability of the lugs in terms of allowable forces which supposes an approach opposite to the one described above. The Neuber hyperbola passing via the coordinate point (εs;σR) cuts the straight line of slope E at coordinate point σ e max (σe max; ). E σ e max Therefore, σR . εs = σe max . E σe max = Kte . σnom adm

Now

The elastic-plastic overstress factor Kσ is defined as:

That is here:

Kσ =

σR σ nom adm

Þ Kσ = Kte .

Þ σnom adm =

Kσ =

σ max σ nom

σR Kσ

σR E . εs

The value of Kσ only has a meaning if Kσ ≥ 1. Otherwise, we consider that Kσ = 1.

Finally, the force bringing the real stress to the maximum allowable value ( σmax = σR) is: Ptu =

At . σR Kσ

V1-7 3.1.4 Calculating εs εs is the strain corresponding to the start the striction, i.e. the unstable state where strain continues at a point without it being necessary to increase the total force F. We then obtain σ = σR. εs is not conventional material data. Nonetheless, it is possible to calculate εs from "n", the Ramberg & Osgood equation factor: εs ≈ 1 ≤ e (%) minimum elongation at break for the material n

If "n" is not material data, solve the following equation: æσ ö n = 500 .ç 0.2 ÷ è σR ø

n



page V1-7•3-3

SSM V1-7 • LUGS

Page Intentionally Left Blank



Issue 1

SSM V1-7 • LUGS V1-7 3.2 SHEAR/BEARING BREAKING STRESS The allowable shear/bearing force at ultimate load is: Pbru = Kbr . Abr . Ftux

where

Ftux =

Fbru 1.9

The factor Kbr depends on ratios a/D and D/t. It is determined using the graph on Figure V1-7.3.2-1.

Figure V1-7 .3.2-1: Determining factor Kbr On the Figure V1-7 .3.2-1 - curve A is the upper limit to be taken into account for all forged aluminium alloy parts when the long transverse direction coincides with direction "C" on the drawing of the lug, - curve B is the upper limit to be taken into account for all aluminium alloy forgings, bars and sheets when the short transverse direction coincides with direction "C" on the drawing of the lug and for die forged parts when the lug contains the jointing plane in a direction approximately perpendicular to direction "C".

In addition to the limits of curves A and B, the value of Kbr will be limited to 2.0 for all aluminium alloy forgings, bars and sheets of a thickness equal to or greater than 12.7 mm.



page V1-7•3-5

SSM V1-7 • LUGS

Page Intentionally Left Blank


Allowable axial force

Issue 1

SSM V1-7 • LUGS V1-7 3.3 SPECIAL CASE CONCERNING EYE ENDS Considering their specific geometry, eye ends are mainly subject to normal circumferential stresses. On Figure V1-7 .3.3-1: - point ② is the RH point from which the load transmitted in the ring is pure tension (normal force), - point ③ is the point where the ring is clamped in the remainder of the part. C ②

① β

③

ϕ Ψ

F r

Figure V1-7 .3.3-1: Eye end Experience proves that the critical section of eye ends is located at ② or at ③. The first step consists in determining the position of ②, knowing the position of ③ and according to a certain number of geometrical parameters of the ring. This calculation requires the use of the graph on Figure V1-7 .3.3-2. The expression of factor K required to read the graph is: K=

C.I A.r3

π . r . (90° + β) 180° I: moment of inertia of the section at ① in relation to its own centre of gravity A: area of the section at ① r: radius of the neutral fibre of the ring

where: C: circular distance (refer to Figure V1-7 .3.3-1)


C=


page V1-7•3-7

SSM V1-7 • LUGS

Figure V1-7 .3.3-2: Determining angle ϕ



Issue 1

SSM V1-7 • LUGS By writing the static equilibrium of the ring, the normal load at② is therefore: N2 =

F 2 . cos ϕ

By calling A2 the section area at ②, the corresponding normal stress is: σ2 =

N2 F = A 2 2 A 2 . cos ϕ

At ultimate load, we will obtain: σ2 = σR

Let Ptu2 be the allowable tensile force at ultimate load conditioned by the static strength at point ②. We then obtain from the previous equality: σR =

Ptu2 2 A 2 . cos ϕ

Hence finally: Ptu2 = 2A2 . cos ϕ . σR And at limit load: Pty2 = 2A2 . cos ϕ . σ0.2 The internal load matrix at ③ is deduced from the value calculated at ②: - normal force:

N3 = N2 . cos Ψ

- shear force:

T3 = N2 . sin Ψ

- bending moment:

M3 = N2 . r . (1 - cos Ψ)

The maximum normal stress at ③ is then:

where:

σ3 =

N 3 M 3 . v 3 N 2 . cos ψ N 2 . r . (1 − cos ψ ) . v 3 + = + A3 I3 A3 I3

σ3 =

F 2 cos ϕ A3: v3: I3:

æ cos ψ r . (1 − cos ψ ) . v 3 ö + .ç ÷ I3 è A3 ø

section area at ③ distance between centre of gravity of section at ③ and extreme fibre moment of inertia of section at ③ in relation to its own centre of gravity

As at ②, at ultimate load, we obtain: σ3 = σR



page V1-7•3-9

SSM V1-7 • LUGS Let Ptu31 be the allowable tensile force at ultimate load conditioned by the static strength under normal stress at point ③. We then obtain from the previous equality: Ptu31 æ cos ψ r . (1 − cos ψ ) . v 3 ö + .ç ÷ 2 cos ϕ è A 3 I3 ø

σR =

Hence finally: Ptu31 =

2σ R . cos ϕ æ cos ψ r . (1 − cos ψ ) . v 3 ö + ç ÷ I3 è A3 ø

Remark: The calculation of allowable force Ptu31 does not take into account the plasticization of the extreme fibres of the section which would "increase" the apparent bending limit. Thus, Ptu31, calculated on the basis of σR, is underestimated and therefore conservative.

Also at limit load: Pty3 =

2σ 0.2 . cos ϕ æ cos ψ r . (1 − cos ψ ) . v 3 ö + ç ÷ I3 è A3 ø

The shear stress at ③ at ultimate load is written: τ3 =

T3 N . sin ψ F .sin ψ = 2 = A3 A3 2 A 3 . cos ϕ

At ultimate load, we obtain: τ3 = τadm Let Ptu32 be the allowable tensile force at ultimate load conditioned by the static strength under tangential stress at point ③. We then obtain from the previous equality: τadm = Ptu32 Hence finally: Ptu32 =

2 A 3 . τ adm . cos ϕ sin ψ

Finally, the overall allowable tensile force at ultimate load is: Ptu = min(Ptu2;Ptu31;Ptu32) At limit load: Pty = min(Pty2;Pty3)



Issue 1

SSM V1-7 • LUGS V1-7 3.4 BLOCK DIAGRAMS V1-7 3.4.1 Solid lugs

E AXIAL LOADING

yes

no

Allowable shear - bearing force at ultimate load: Pbru = Kbr . Abr . Ftux F ( e / D = 2) Ftux = bru 1.9

n known? Solve:

æσ ö n = 500 . ç 0.2 ÷ è σR ø

n

+ graph on Figure V1-7 .3.2-1

Striction strain: εs =

S

Determine Kt from graphs in FATIGUE MANUAL MTS 005

1 ≤ e(%) n

Kte =

Kσ = Kte .

yes

S

E

W −D . Kt W

N

σR E . εs no

Kσ ≥ 1 ?

T

Kσ = 1

Ptu =

At . σR Kσ

I

Allowable axial force at ultimate load: Pu = min (Ptu; Pbru)

Ftyx = 0.593.Fbry (e / D = 2)

Pu ≥ 1,.5 A br . Ftux

yes Ky = 1.3625 – 0.25

Pu A br Ftux

A

no Ky = 1.1

L

Allowable axial force at limit load: Ftyx Py = K y . . Pu Ftux Issue 1 © Copyright AEROSPATIALE 1998


page V1-7•3-11

SSM V1-7 • LUGS V1-7 3.4.2 Eye ends

E

AXIAL LOADING

Evaluation of position of point ③ (angle β)

S S

Circular distance:

Determining angle ϕ from K =

C=

π . r . (90° + β) 180°

C.I and graph on Figure V1-7 .3.3-2 A . r3

E

Allowable tensile force at U.L. conditioned by the static strength at point ②: Ptu2 = 2A2 . cos ϕ . σR

Allowable tensile force at L.L. conditioned by the static strength at point ②: Pty2 = 2A2 . cos ϕ . σ0.2

N

Allowable tensile force at ultimate load conditioned by the static strength under normal stress at point ③: 2σ R . cos ϕ Ptu31 = æ cos ψ r . (1 − cos ψ ) . v 3 ö + ç ÷ I3 è A3 ø

Allowable tensile force at limit load conditioned by the static strength at point ③: 2σ 0.2 . cos ϕ Pty3 =

T I A

æ cos ψ r .(1 − cos ψ ). v3 ö + ç ÷ I3 è A3 ø

Allowable tensile force at ultimate load conditioned by the static strength under tangential stress at point ③: 2 A 3 . τ adm . cos ϕ Ptu32 = sin ψ

Allowable tensile force at ultimate load: Ptu = min (Ptu2; Ptu31; Ptu32)

Allowable tensile force at limit load: Pty = min (Pty2; Pty3)

L page V1-7•3-12 © Copyright AEROSPATIALE 1998


Issue 1


ALLOWABLE TRANSVERSE FORCE

In spite of the combination of several distinct failure modes, the transverse capability of the lug is expressed by a single allowable force. The allowable transverse force at ultimate load is: Ptru = Ktru . Abr . σR The allowable transverse force at limit load is: Ptry = Ktry . Abr . σ0.2 Factors Ktru and Ktry are obtained from the graph, Figures V1-7 .4-2 to V1-7 .4-11. However, before using these graphs you must calculate an equivalent section Aav, which takes the various failure modes in the transverse direction into account. The definition is given in Figure V1-7 .4-1.

R

A4

A4

A3

45°

45°

45°

A1 A1

45°

A2

CONVENTIONAL LUG

Transverse direction = Direction of force

EYE END

Figure V1-7 .4-1: Definition of the equivalent cross section Aav We can write:

A av =

6 3 1 1 1 + + + A1 A 2 A 3 A 4

- A2 is measured at the hole in the transverse force direction - with a conventional lug: A1 and A4 are measured parallel to the transverse direction along lines offset by 45° (refer to Figure V1-7 .4-1). - with an eye end: A1 and A4 are measured perpendicularly to the circular neutral line of the lug (refer to Figure V1-7 .4-1). - A3 is the smallest radial cross section around the hole.

A av is used which represents the shape of A br lug. Note that unlike Ktru, Ktry does not depend on the material concerned.

To read the graph and therefore determine Ktru and Ktry, parameter


Allowable transverse force

page V1-7•4-1

SSM V1-7 • LUGS

Ktru or Ktry

Figure V1-7 .4-2: Determining factors Ktru and Ktry Interpretation of the curves above makes reference to Table V1-7 .4-3. Note that these curves only apply if the short transverse direction is not in the plane of the lug. Curve 1 2 3 4 5 6 7 8 9

10 11 12

Material concerned steel 25CD4S and 25CD45 86 daN/mm² steel 25CD4S and 25CD45 103 daN/mm² Ktry for all materials steel 25CD4S and 25CD45 124 daN/mm² 2024-T3 and -T4 plates, thickness ≤ 12.7 mm 2014-T6 and 7075-T6 plates, thickness ≤ 12.7 mm 2024-T3 and -T4 plates, thickness > 12.7 mm, 2024-T4 bar cast 356-T6 2024-T6 and 7075-T6 plates, thickness > 12.7 mm, ≤ 25.4 mm 7075-T6 extrusions 2014-T6 forgings, cross section ≤ 23224 mm² 2014-T6 and 7075-T6 die forged parts 2024-T6 plates 2024-T4 and 2024-T42 extrusions 2014-T6 and 7075-T6 plates, thickness > 25.4 mm 7075-T6 forgings, cross section ≤ 10322 mm² 7075-T6 forgings, cross section > 10322 mm² 2014-T6 forgings, cross section > 23224 mm² Table V1-7 .4-3: References for curve .V1-7.4-2



Issue 1

SSM V1-7 • LUGS 1.7

1

1.6

2

1.5 1.4 1.3

3

1.2

4

1.1 1.0 Ktru or Ktry

0.9 0.8

5

0.7 0.6 0.5 A

0.4 0.3

6

0.2 0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 A ay A br

Figure V1-7 .4-4: Determining factors Ktru and Ktry Interpretation of the curves above makes reference to table V1-7 .4-5. Curve 1 2 3 4 5 6

Material concerned steel < 87 daN/mm² steel 105 daN/mm² Ktry for all materials steel 126 daN/mm² steel 140 daN/mm² steel 182 daN/mm² Table V1-7 .4-5: References for curve V1-7 .4-4



page V1-7•4-3

SSM V1-7 • LUGS 1.0

1

.8

A Ktru

2 .6 4 3

5 .4

6 7 A

.2

0 0

.2

.4

.6

A ay

.8

1.0

1.2

1.4

A br Figure V1-7 .4-6: Determining factors Ktru Interpretation of the curves above makes reference to table V1-7 .4-7. Curve 1

2

3 4

5

6 7

Material concerned Ti-6A1-4V die forged annealed (LT), thickness ≤ 127 mm Ti-6A1-4V forged annealed (LT), cross section ≤ 10323 mm² Ti-6A1-4V forged (L, LT), thickness ≤ 51 mm Ti-6A1-4V forged annealed (LT), cross section > 10323 mm² Ti-6A1-4V die forged (L), thickness ≤ 127 mm Ti-6A1-4V die forged (LT), thickness ≤ 25 mm Ti-6A1-4V forged (LT), thickness > 51 mm, ≤ 76 mm Ti-6A1-4V die forged (LT), thickness > 25 mm, ≤ 76 mm Ti-6A1-6V-2Sn plate annealed (LT), thickness ≤ 51 mm Ti-6A1-6V-2Sn die forged annealed (ST), thickness ≤ 51 mm Ti-6A1-6V-2Sn forged annealed (LT), thickness ≤ 51 mm Ti-6A1-6V-2Sn plate annealed (LT), thickness > 51 mm, ≤ 101 mm Ti-6A1-6V-2Sn die forged annealed (ST), thickness > 51 mm, ≤ 101 mm Ti-6A1-6V-2Sn forged annealed (LT), thickness > 51 mm, ≤ 101 mm Ti-6A1-6V-2Sn die forged (L) Ti-6A1-6V-2Sn forged (L, LT), thickness ≤ 101 mm Ti-6A1-6V-2Sn die forged (LT) Ti-6A1-6V-2Sn forged (LT), thickness > 101 mm Table V1-7 .4-7: References for curve V1-7 .4-6



Issue 1

SSM V1-7 • LUGS

1.2

2

1.0

Ktru

.8 1 A .6

.4

3 .2 A 0 0

.2

.4

.6

.8

1.0

1.2

1.4

A ay A br

Figure V1-7 .4-8: Determining factors Ktru Interpretation of the curves above makes reference to table V1-7 .4-9. Curve 1 2 3

Material concerned steel 300M with e% ≥ 6 low alloy steels with Ftu ≤ 210 daN/mm² (except 300M) low alloy steels with 3 ≤ e% ≤ 6 Table V1-7 .4-9: References for curve V1-7 .4-8



page V1-7•4-5

SSM V1-7 • LUGS

1 1.4 2 1.3 1.2 3

1.1 1.0

4

.9 Ktru A

.8 .7

5

.6 .5 .4 .3 A .2 .1 0 0

.1

.2

.3

.4

.5

.6

.7

.8

.9

1.0

1.1

1.2

1.3

1.4

A av A br Figure V1-7 .4-10: Determining factors Ktru and Ktry Interpretation of the curves above makes reference to table V1-7 .4-11. Curve 1 2 3 4 5

Material concerned steel 15-5PH H1150 with Ftu = 95 daN/mm² and e% = 11 (T) steel 15-5PH H1100 with Ftu = 98 daN/mm² and e% = 10 (T) Inconel steel 718 B&F Sta with Ftu = 126 daN/mm² and e% = 10 (LT) steel 15-5PH H1025 with Ftu = 109 daN/mm² and e% = 8 (T) Inconel steel 718 B&F Sta with Ftu = 126 daN/mm² and e% = 6 (T) Table V1-7 .4-11: References for curve V1-7 .4-10



Issue 1

SSM V1-7 • LUGS If the material is not in the list given in Tables V1-7 .4-3, V1-7 .4-5, V1-7 .4-7, V1-7 .4-9 or V1-7 .4-11, an analogy may be made based on: - the same type of finished product (plate, extrusion, bar, forging, etc.), - a similar ductility (in particular, expressed by the elongation at break), - a similar σR/σ0.2 ratio. If no analogy can be found, a finite element analysis may be used in compliance with the following rules: - take into account any possible anisotropy due to the manufacturing process and correct orientation of the model in relation to the longitudinal, long transverse and short transverse directions (the results of the study above prove that this parameter has an affect on the transverse strength). If the direction of these fibres is not known, the long transverse direction may be placed unfavourably parallel to the transverse direction. Remember that the plane of the lug must not contain the short transverse direction, - nonlinear calculation in the plastic domain at ultimate load. Globally, if the dimensions of the hole are small compared with the lug, this being quantified by the ratio Aav/Abr, the local transverse failure phenomena become negligible in relation to the overall bending strength of the lug. This transition is approximately indicated by curve A on graphs V1-7 .4-2 to V1-7 .4-11. If Ktru is below this curve, it is advisable to carry out a fixed end beam calculation using a cross section as shown on Figure V1-7 .4-12. Only take the lower section of the lug into account. Depending on the shape and the material of the lug, Ktru may be very conservative. Ptru is in no case lower than the ultimate bending load of the ligament under the transverse load.

F

Figure V1-7 .4-12: Approximate portion of the lug acting as a fixed end beam The approach consists in calculating the inertia and bending moment in the most critical cross section, moving away from the hole (force application point - refer to Figure V1-7 .4-13).

ji

F

n

i

21

Figure V1-7 .4-13: Calculation of bending moment in lug flange For a critical cross section "i", the maximum bending stress is:

σi =

M f i . vi Ii

=

F . ji . vi Ii



page V1-7•4-7

SSM V1-7 • LUGS where: Mfi: bending moment due to F in cross section "i" vi: distance between the centre of gravity and the extreme fibre of cross section "i" ji: lever arm of Mfi Ii: moment of inertia of cross section "i" in relation to its own centre of gravity Taking as condition σiMAX = σR , the allowable force at U.L. is therefore:

æ I ö Ptru = σR . min ç i ÷ è ji . vi ø i =1..n And at L.L.

æ I ö Ptry = σ0.2 . min ç i ÷ è ji . vi ø i =1..n



Issue 1

SSM V1-7 • LUGS TRANSVERSE LOADING yes

E

no Material referenced? yes

no

Analogy?

Finite element analysis Determine A1, A2, A3 and A4, then Aav

S

R

A4

A4

A3

45°

45°

where Aav

6 3 1 1 1 + + + A1 A 2 A 3 A 4

45°

A1 A1

45°

S

A2 Transverse direction

CONVENTIONAL LUG

EYE END

E

Interpretation of graphs: V1-7 .4-2 to V1-7 .4-11

N yes

no

Fixed end beam?

T

Ji

F

n

i

Determination of Ktru and Ktry

I

21

æ I ö Allowable force at U.L. Ptru = σR . min ç i ÷ è ji . vi ø i =1..n æ I ö Allowable force at L.L. Ptry = σ0.2 . min ç i ÷ è ji . vi ø i =1..n

Allowable force at U.L. Ptru = Ktru . Abr . σR Allowable force at L.L. Ptry = Ktry . Abr . σ0.2

A L



page V1-7•4-9

SSM V1-7 • LUGS




Issue 1


E

STRENGTH UNDER AN OBLIQUE FORCE

In order to check the strength of the lug under an oblique force, the force must be projected in the transverse and axial directions as shown on Figure V1-7 .5-1.

Transverse direction

S

Fa

Axial direction

θ F Ftr

S E N

Figure V1-7 .5-1: Lug under an oblique force Notation: Fa = F . cos θ the axial component of the force (at U.L.) Ftr = F . sin θ the transverse component of the force (at U.L.) We consider that the force at limit load is obtained by dividing the force at U.L. by 1.5: thus, we obtain: F Fa = . cos θ the axial component of the force (at L.L.) 1.5 F Ftr = . sin θ the transverse component of the force (at L.L.) 1.5

In all cases, the interaction equation between the axial and transverse loads is:

T I A

R 1a.6 + R1tr.6 = 1

(document reference (5) §D.1.8)

where: Ra =

Fa (at ultimate load) Pu

Ra =

Fa (at limit load) Pbry

Rtr =

Ftr (at ultimate load) Ptru

Rtr =

Ftr (at limit load) Ptry

We can therefore calculate a "reserve factor" for this combined loading using the interaction equation. Finally, we obtain: R.F. =

(R

1 1.6 a

+ R 1tr.6

)

0 .625

L Issue 1 © Copyright AEROSPATIALE 1998

Strength under an oblique load

page V1-7•5-1

SSM V1-7 • LUGS OBLIQUE LOADING

Ultimate load calculation

Limit load calculation

Axial and transverse projection of F: Fa = F . cos θ Ftr = F . sin θ

Axial and transverse projection of F: Fa = F/1.5 . cos θ Ftr = F/1.5 . sin θ

Comparative ratios: F Ra = a Pu F Rtr = tr Ptru

Comparative ratios: F Ra = a Pbry Rtr =

Ftr Ptry

Calculation of "reserve factor" 1 R.F. = 0 .625 1.6 R a + R 1tr.6

(


)

Strength under an oblique load

Issue 1


ALLOWABLE BEARING IN BUSH

V1-7 6.1 ALLOWABLE BEARING IN LUGS The bearing cross section involved (Abrb) will be the minimum area between: - the bearing area between the pin and the bush, - the bearing area between the bush and the lug. The latter may be smaller if, for example, the bush includes an outside chamfer. The allowable limit force for this type of damage is: P' bry = 1.85 . σc0.2 . Abrb with: σc0.2: 0.2% compression yield strength for the bush material P'bry will be directly compared with the oblique force at limit load.


Allowable bearing

page V1-7•6-1

SSM V1-7 • LUGS



Allowable bearing

Issue 1

SSM V1-7 • LUGS

V1-7 7

PIN STRESSING

V1-7 7.1 PIN SHEAR The shear stress in the pin can be written: τ=

F As

As

:

where

d:

sheared surface area of the pin, such that A s =

π d2 . 4

pin diameter.

The calculation of the corresponding allowable forces is given below. For a male / female part assembly: π d2 4 (double shear)

Psu = τ adm × 2 ×

That is,

Psu =

Where: τ adm :

allowable pure shear stress.

Remark:

Shear is only calculated at ultimate load.

π d 2 τ adm 2

V1-7 7.2 PIN BENDING The bending of the lug pins must be studied both at the limit load and at the ultimate load. Indeed, permanent sets are unacceptable at limit load as they hinder or even prevent the disassembly of the junction. The allowable bending forces at ultimate load and at limit load are called Pfu and Pfy respectively. The maximum bending stress will be located at the extreme fibre of the lug pin. By calling σ app the stress applied (due to bending), we obtain: σapp =

Mf ⋅ v Mf ⋅ d b ⋅ F ⋅ d = = I 2I 4I

V1-7 7.2.1 Calculation of lever arm "b" Conservatively, the calculation of the lever arm to determine the bending moment is often based on the assumption that the contact pressure between the pin and the bores is constant over the length of the pin. In actual fact, due to bending strain, the resisting force exerted by the male portion of the lug is concentrated close to the outer faces and the resisting force exerted by the female portion near to the inner faces (see figure V1-7 .7.2.1-1). This phenomenon, called "peaking", reduces the lever arm which reduces the bending moment in the pin. Moreover, we have demonstrated by tests (document ref. (3)) that the "peaking" does not reduce the strength of the lugs.

Issue 1 ã Copyright AEROSPATIALE 1998

Pin stressing

page V1-7•7− −1

SSM V1-7 • LUGS t1 g

F/2

F/2

F/2

F/2

Female part

b

b

Inactive portions of the thicknesses

Male part F

F

t2 UNIFORM DISTRIBUTION THICKNESSES

OVER

ALL

DISTRIBUTION TAKING PEAKING INTO ACCOUNT

Figure V1-7 .7.2.1-1: Distribution of the contact pressure between the pin and the bores In the first case shown on figure V1-7 .7.2.1-1, the lever arm is: b=

t1 t 2 + +g 2 4

where: t1: t2:

thickness of one of the sides of the female portion of the lug thickness of the male portion of the lug

g:

clearance between the male and female parts

Taking "peaking" into account gives a new expression for the lever arm: t ö æt b = γ ç 1 + 2 ÷+g è2 4ø

For a "male - male" assembly, the lever arm can be written:

b=

t1 t 2 + +g 2 2

Determining factor γ: First of all, the

Dù é êa − 2 ú û. quantity r = ë t2

must be determined. "a" is then measured in the direction of the

applied force whether the load is purely axial, purely transverse or oblique. It is also necessary to evaluate the ratio

Padm A br ⋅ Ftux

called x.

Padm is

defined as being the allowable force by writing, in the general case of an oblique force:

Padm =

(

F

R1a,6

+ R1tr,6

)

0,625

We then obtain γ from the graph on figure V1-7 .7.2.1-2 or from the following equations: ì ï ï ï ï í ï ï ï ï î

r ü ï 0,55 ï x ý Þ γ= 2 r ï x ≥ 1+ 1− ï 0,55 þ x ≤ 1− 1−

æ ö ç1 − 1 − r ≤ x ≤ 1 + 1 − r ÷ Þ γ = 1 − r ç 0,55 0,55 ÷ø 1,1 x è

page V1-7•7− −2 ã Copyright AEROSPATIALE 1998

Pin stressing

Issue 1

SSM V1-7 • LUGS

Figure V1-7 .7.2.1-2: Determining peaking factor γ

V1-7 7.2.2 Calculating bending modulus We can see that an item subjected to bending is capable of supporting a load greater than the calculated load assuming a linear distribution of the stresses over the height of the section. This is due to the fact that the extreme fibres plasticize which modifies the normal stress profile. Therefore, the real stress in these fibres is lower than that obtained by a simple linear calculation (refer to Figure V1-7 .7.2.2-1).

z

zσ MAX

z

σMAX x

x

y

σMAX

x

STRESS CALCULATED BY LINEAR METHOD

σMAX REAL STRESS

Figure V1-7 .7.2.2-1: Bending stress On the figure above, σMAX is the stress applied to the extreme fibre. The Cozzone method is based on an approximation of the real stress profile. The real stress profile shown on Figure V1-7 .7.2.2-1 is replaced by a trapezoidal profile as shown on Figure V1-7 .7.2.2-2.

σMAX

z

σ

σMAX

Equivalent profile

σ0

Material curve

ε

σ0

σ0

x

σMAX

εMAX

Figure V1-7 .7.2.2-2: Cozzone's equivalent stress profile


Pin stressing

page V1-7•7− −3

SSM V1-7 • LUGS Stress σ 0 is a fictitious stress that is supposed to exist at the neutral fibre of the pin or at null strain which is the same thing here. We determine σ 0 by stating that σ 0 does not theoretically depend on the shape of the section. The Cozzone method enables a fictitious allowable bending stress called "bending modulus" to be calculated. This stress can be compared with the maximum stress calculated by a linear method.

σb = σMAX + σ0 . (k - 1)

The general expression of the bending modulus is:

Remarks: ① The calculation of the bending modulus is based on the real desired maximum stress level, that is generally σ MAX = σ R the ultimate strength for a calculation at ultimate load. At limit load, the stress level must remain lower than the yield strength.σb = σMAX + σ0 . (k - 1) ② The bending modulus also depends on the geometrical characteristics of the section by means of the term "k". "k" is determined from figure V1-7 .7.2.2-3 or from the following equation. σb = σMAX + σ0 . (k - 1) æd 1 − çç i 16 è d k= ⋅ 3π æd 1 − çç i è d

Where: d: di :

ö ÷÷ ø

3

ö ÷÷ ø

4

pin outside diameter. pin inside diameter.

1.85 1.75 1.65 k

1.55 1.45 1.35 1.25 0.0

0.2

0.4

0.6 0.8 di/d

1.0

1.2

Figure V1-7 .7.2.2-3: Determining "k"


Pin stressing

Issue 1

SSM V1-7 • LUGS The stress σ 0 is determined from the various graphs shown on figures V1-7 .7.2.2-5 to -19. If none of the curves corresponds to the material used, σ 0 must be calculated knowing that this stress does not theoretically depend on the shape of the section. That which is valid for a rectangular section is also valid for all other sections. We have checked that the method given below to calculate the bending modulus σ 0 gives the same results as the graphs shown on figures V1-7 .7.2.2-5 to -19. σ 0 corresponding to σ R is calculated; this stress level is obtained on the extreme fibre of the pin. Considering this

as pure bending and that the material curve is perfectly symmetrical in tension-compression, we assume that stress is null at the neutral fibre of the pin. Consequently, a second x-axis can be plotted corresponding to the dimension between the neutral fibre and a fibre given on the material curve (refer to figure V1-7 .7.2.2-4).

σ σR σ

z Zmax = h/2

σ(z)

z

0 0

b

εs z

ε

z zmax = d/2

Figure V1-7 .7.2.2-4: Real stress profile

Thus, the bending moment can be written: h/2

M=2b

ò z ⋅ σ(z) ⋅ dz 0

Where:

h ε ì ïz = 2 × ε ï M Plane bending: the straight sections remain plane. ε is linear with z. í ïdz = h dε ïî 2ε M

Also, ε can be written: ε=

æ σ ö σ ÷ + 0,002ç ç σ0, 2 ÷ E è ø

n

æ σ 0, 2 ö ÷ ÷ è σR ø

n

Where: n = 500çç


Pin stressing

page V1-7•7− −5

SSM V1-7 • LUGS We deduce the expressions of ε and ε max : σ 1æ σ ε = + çç E n è σR

n

ö ÷ = εe + ε p ÷ ø n

ε max =

σMAX 1 æ σMAX ö ÷ = εe + çç + ε p MAX MAX E n è σR ÷ø

M⋅v 6 = ⋅M I b ⋅ h2

σb =

However Also:

ìσb = (k − 1) ⋅ σ0 + σMAX ï í h⋅W 3 = ïk = I 2 î

æ σb ö σ0 = 2çç − 1÷÷ σMAX σ è MAX ø

Þ

After resolution, we obtain: σ0

σMAX

=

2 2(n − 1) æ ε p MAX ö æç æ 2n + 1 ö æç εe MAX ÷ ⋅ ç ⋅ çç ÷⋅ 2n + 1 è εMAX ÷ø ç è n + 2 ø çè εp MAX è

ö ö ÷ + 1÷ ÷ ÷ ø ø

This expression can also be written: σ0 σ MAX

æ n − 1 ö æç ε p MAX ö÷ æç æ n − 1 ö æç ε p MAX ö÷ ö÷ = 2⋅ç ⋅ 1− ç ÷⋅ ÷⋅ è n + 2 ø çè ε MAX ÷ø çè è 2n + 1 ø çè ε MAX ÷ø ÷ø

Where: σ MAX

æ

ö

1 σ = çç MAX ÷÷

ε p MAX ε MAX

n

maximum allowable stress n

n è σR ø σ = MAX + ε p MAX E æ σ 0, 2 ö ÷ ÷ è σR ø

maximum allowable plastic strain maximum allowable strain

n

=500çç

(numerical resolution)

Specific cases: σR 1 + E n σ 0, 2 = 0,2% ; ε MAX = + 0,2% E

①

(σMAX = σR )

Þ ε p MAX =

②

(σMAX = σ0,2 )

Þ ε p MAX

1 n


; ε MAX =

Pin stressing

Issue 1

SSM V1-7 • LUGS

Figure V1-7 .7.2.2-5: Curves σMAX = f(ε) and σ0 = f(ε) - 2024-T3 clad sheet (0.25 mm < thickness < 1.6 mm)

Figure V1-7 .7.2.2-6: Curves σMAX = f(ε) and σ0 = f(ε) - 2024-T6 clad sheet (thickness < 1.6 mm)

Figure V1-7 .7.2.2-7: Curves σMAX = f(ε) and σ0 = f(ε) - 2024-T4 clad sheet (6.3 mm < thickness < 12.7 mm)


Pin stressing

page V1-7•7− −7

SSM V1-7 • LUGS


Figure V1-7 .7.2.2-9: Curves σMAX = f(ε) and σ0 = f(ε) - 2024-T3 plate and sheet (thickness ≤ 6.35 mm)

Figure V1-7 .7.2.2-10: Curves σMAX = f(ε) and σ0 = f(ε) - 2024-T3 & -T4 plate and sheet (thickness ≤ 12.7 mm)



Pin stressing

Issue 1

SSM V1-7 • LUGS

Figure V1-7 .7.2.2-12: Curves σMAX = f(ε) and σ0 = f(ε) - 7075-T6 sheet (thickness < 1 mm)

Figure V1-7 .7.2.2-13: Curves σMAX = f(ε) and σ0 = f(ε) - 7075-T6 extrusion (thickness < 6.35 mm)

Figure V1-7 .7.2.2-14: Curves σMAX = f(ε) and σ0 = f(ε) - 7075-T6 forged part (thickness < 10322 mm²)

Figure V1-7 .7.2.2-15: Curves σMAX = f(ε) and σ0 = f(ε) - AZ61A forged magnesium alloy part (direction L)


Pin stressing

page V1-7•7-9

SSM V1-7 • LUGS

Figure V1-7 .7.2.2-16: Curves σMAX = f(ε) and σ0 = f(ε) - T-A6V

Figure V1-7 .7.2.2-17: Curves σMAX = f(ε) and σ0 = f(ε) - T-M4A4

Figure V1-7 .7.2.2-18: Curves σMAX = f(ε) and σ0 = f(ε) - Z6 CNU 17.04 bar and forged part

Figure V1-7 .7.2.2-19: Curves σMAX = f(ε) and σ0 = f(ε) - 7079-T6 die forged part (L direction) (thickness ≤ 152.4 mm) page V1-7•7-10 © Copyright AEROSPATIALE 1998

Pin stressing

Issue 1

SSM V1-7 • LUGS V1-7 7.2.3 Allowable bending force o calculate Pfu and Pfy , we make the linear bending stress expression equal to the Cozzone expression. Thus,

generally: F⋅b⋅d = σMAX + σ0 ⋅ (k − 1) 4Ι

At ultimate load, we obtain: σ MAX = σR Hence: Pfu ⋅ b ⋅ d = σR + σ0 ⋅ (k − 1) 4Ι

Þ

Pfu =

4Ι [σR + σ0 ⋅ (k − 1)] b⋅d

Where: I: b: d: σR: σo: k:

Pin moment of inertia bending moment lever arm (refer to §V1-7 .7.2.1) pin diameter ultimate strength Cozzone fictitious stress at neutral fibre (refer to §V1-7 7.2 2) σ 0 factor (refer to §V1-7 .7.2.2).

At limit load, the stress levels must remain within the elastic range which means that σ 0 =0 and σ b =σ MAX =σ0, 2 . Therefore, the allowable force at limit load will be simply: Pfy =

4 Ι ⋅ σ 0, 2 b⋅d

Where: σ0, 2 : 0.2% tensile yield strength.


Pin stressing

page V1-7•7-11

SSM V1-7 • LUGS PIN BENDING

E

Dù é êa − 2 ú Padm ë û Determine the dimensionless quantity r = and ratios: (ultimate load) and t2 A br ⋅ σR

Padm (limit load) where Padm = A br ⋅ σ 0, 2

(R

F 1,6 a

+ R 1tr,6

)

0,625

S

Determine γ at U.L. and L.L. with the graph on figure V1-7 .7.2.1-2

t ö æt b = γ ⋅ çç 1 + 2 ÷÷ + g 4 ø è2 t1 t 2 b= + +g 2 2

Calculate the lever arm at U.L. and L.L.:

("male / male" assembly)

Calculation at limit load

Calculation at ultimate load

( D) Calculate: k = 1 − (d ) D 1− d

Material known?

E

3 4

yes

S

("male / female" assembly)

⋅

N

16 3π

no yes

n known?

Determine σ 0 from the graphs on figures V1-7 .7.2.2-5 to V1-7 .7.2.2-19

T

no

Solve: æ σ 0, 2 n = 500çç è σR

ö ÷ ÷ ø

n

I

Calculation of stress at neutral fibre: æ n − 1 ö æç ε p MAX ö÷ æç æ n − 1 ö æç ε p MAX ö÷ ö÷ = 2ç ⋅ 1−ç ÷⋅ ÷⋅ è n + 2 ø çè ε MAX ÷ø çè è 2n + 1 ø çè ε MAX ÷ø ÷ø

σ0 σ MAX

A

where: n

ε p MAX

ö 1æσ σ = çç MAX ÷÷ ;ε MAX = MAX + ε p MAX n è σR ø E

Allowable bending force at U.L.: Pfu

4Ι = [σ R + σ 0 ⋅ (k − 1)] b⋅d


Pin stressing

L

Allowable bending force at L.L.: Pfy =

4 Ι ⋅ σ 0, 2 b⋅d

Issue 1

SSM V1-7 • LUGS V1- 8

NUMERICAL EXAMPLE

V1-7 8.1 BASIC DATA t2 = 18 mm

D = 18 mm R 20 mm

g = 0.6 mm t1 = 10 mm

Figure V1-7.9.1-1: Example The male and female parts are taken from a thick 2024 T351 plate. The characteristics taken into account are: σR = 44 daN/mm² σ0.2 = 30.5 daN/mm². E1 = E2 = 7380 daN/mm² Fbru (e/D=2) = 82 daN/mm² Fbry (e/D=2) = 62.5 daN/mm² n = 9 e(%) = 6 The pin is made of TA6V; we consider that the pin diameter is 18 mm. The characteristics taken into account are: σR = 90 daN/mm² n = 40 Ea = 11030 daN/mm² σ0.2 = 87 daN/mm² τadm = 52 daN/mm² According to the geometrical information above, we have: W = 40 mm L = a = 20 mm R = 20 mm D = 18 mm d ≈ 18 mm t1 = 10 mm t2 = 18 mm g = 0.6 mm The critical surface areas considered are: A t1 = (W − D ) ⋅ t1 = (40 − 18) × 10

A t1 = 220 mm 2

A br1 = D ⋅ t1 = 18 × 10

A br1 = 180 mm 2

A t 2 = (W − D ) ⋅ t 2 = (40 − 18 ) × 18

A t 2 = 396 mm 2

A br 2 = D ⋅ t 2 = 18 × 18

A br 2 = 324 mm 2

Force applied: The components of the oblique force to be transferred at ultimate load are: Fa = 7000 daN Ft = 3000 daN We consider that these forces already take a fitting factor into account (1.15 for instance). Issue 1 ã Copyright AEROSPATIALE 1998

Numerical Example

page V1-7•8-1

SSM V1-7 • LUGS V1-7 8.2 LUG STRESSING Calculating the allowable axial forces Calculating the allowable tensile forces We determine the breaking strain:

1 1 = = 0,11 > e(% ) Þ ε s = e(% ) = 6 % n 9 The elastic overstress factors are determined using the graphs in the AEROSPATIALE FATIGUE MANUAL. Considering the lug geometry and using the notations in the FATIGUE MANUAL, we obtain: KtET = 5.25 Male part: Female part:

G ≈ 1.1 G ≈ 1.2

Therefore: Male part: Female part:

Kt = Kt ET ⋅ G = 5,25 × 1,1 = 5,78 Kt = Kt ET ⋅ G = 5,25 × 1,2 = 6,3

The reference section used in the FATIGUE MANUAL is the gross section whereas it is the net section which is taken into account in the S.S.M. Therefore, the overstress factors to be considered below are as follows: Male part: Female part:

W −D 40 − 18 ⋅ Kt = ⋅ 5,78 = 3,18 W 40 W −D 40 − 18 Kte = ⋅ Kt = ⋅ 6,3 = 3,47 W 40

Kte =

The overstress factors Kσ are then: Male part:

Kσ = K te ⋅

σR 44 = 3,18 × ≈1 E ⋅ εs 7380 × 0,06

Female part:

Kσ = K te ⋅

σR 44 = 3,47 × ≈ 1,09 E ⋅ εs 7380 × 0,06

Thus, the allowable tensile forces at ultimate load are:

Male part:

Ptu =

A t 2 ⋅ σ R 396 × 44 = Kσ 1

Ptu = 17420 daN

Female part:

Ptu = 2 ×

A t1 ⋅ σR 220 × 44 = 2× Kσ 1,09

Ptu = 17760 daN


Numerical Example

Issue 1

SSM V1-7 • LUGS Calculating allowable shear / bearing forces:

For the male and female parts, we obtain:

a = 1,11 D

According to the graphs on figures V1-7 .3.2-1 and V1-7 .3.2-2,

Male part:

Female part:

a ü = 1,11ï D ï ý Þ Kbr ≈ 0,98 D =1 ï ïþ t2 a ü = 1,11ï D ï ý Þ Kbr ≈ 0,98 D = 1,8 ï ïþ t1

Thus, the allowable forces at ultimate load are: Male part:

Pbru = Kbr ⋅ A br 2 ⋅ Ftux = 0,98 × 324 ×

82 1,9

Pbru = 13700 daN

Female part:

Pbru = 2 × Kbr ⋅ A br1 ⋅ Ftux = 2 × 0,98 × 180 ×

82 1,9

Pbru = 15220 daN

Summary of allowable axial forces at ultimate load: Male part:

Pu = min (Ptu ; Pbru) = min (17420 ; 13700) Pu = 13700 daN

Female part:

Pu = min (Ptu ; Pbru) = min (17760 ; 15220) Pu = 15220 daN

Ftyx = 0,593 × Fbry = 0,593 × 62,5 = 37,1 daN / mm²

Male part:

Pu 13700 = = 0,98 < 1,05 82 A br 2 × Ftux 324 × 1,9 Þ K y = 1,1 Pu

Female part:

2 × A br1 × Ftux

=

15220 2 × 180 ×

82 1,9

= 0,98 < 1,05

Þ K y = 1,1 Summary of allowable axial forces at limit load: Ftyx Py = K y × × Pu = 12940 daN Male part: Ftux Female part:

Py = K y ×


Ftyx Ftux

× Pu = 14380 daN

Numerical Example

page V1-7•8-3

SSM V1-7 • LUGS Calculating allowable transverse forces

The geometry of the lugs gives: Male part:

A av =

6 6 = × 18 = 227,3 mm2 3 1 1 1 3 1 1 1 + + + + + + A1 A 2 A 3 A 4 13,64 11 11 13,64

Female part:

A av =

6 6 = × 10 = 126,3 mm2 3 1 1 1 3 1 1 1 + + + + + + A1 A 2 A 3 A 4 13,64 11 11 13,64

According to the graph of figure V1-7 .4-2 (curves 3 and 7), we obtain the following results for both the male and female part: A av A = av ≈ 0,7 Þ K tru = 0,54 and K try = 0,8 A br 2 A br1 Thus, the allowable forces at ultimate load are:

Male part:

Female part:

Ptru = K tru ⋅ A br 2 ⋅ σ R = 0,54 × 324 × 44 Ptru = 7700 daN Ptru = 2 × K tru ⋅ A br1 ⋅ σ R = 2 × 0,54 × 180 × 44 Ptru = 8550 daN

The allowable forces at limit load are: Male part:

Female part:

Ptry = K try ⋅ A br 2 ⋅ σ 0 , 2 = 0,8 × 324 × 30,5 Ptry = 7900 daN Ptry = 2 × K try ⋅ A br1 ⋅ σ 0, 2 = 2 × 0,8 ×180 × 30,5 Ptry = 8780 daN

Calculating the breaking load as if only the ligament subjected to the load was supporting the transverse force would be more conservative still. For this, we retain: Ptru = Ptry .

Summary of allowable transverse forces: At ultimate load:

Male part:

Ptru = 7900 daN

Female part:

Ptru = 8780 daN

At limit load:


Male part:

Ptry = 7900 daN

Female part:

Ptry = 8780 daN

Numerical Example

Issue 1

SSM V1-7 • LUGS Calculation of the lug reserve factors

The reserve factors at ultimate load are: Male part: R.F. =

[

1

]

=

]

=

0, 625 R 1a,6 + R 1tr,6

1 éæ F êç a êçè Pu ë

ö ÷ ÷ ø

1,6

æ F + çç tr è Ptru

ö ÷ ÷ ø

1, 6 ù

0,625

=

1, 6 ù

0,625

=

ú ú û

1 éæ 7000 ö1,6 æ 3000 ö1,6 ù êç ÷ +ç ÷ ú è 7900 ø úû êëè 13700 ø

0, 625

= 1,45

0, 625

= 1,61

Female part: R.F. =

[

1

0, 625 R 1a,6 + R 1tr,6

1 éæ F êç a êçè Pu ë

ö ÷ ÷ ø

1, 6

æ F + çç tr è Ptru

ö ÷ ÷ ø

ú ú û


Let us consider that the limit loads are equal to the ultimate loads divided by 1.5. We then obtain: Fa = 4667 daN Ft = 2000 daN The reserve factors at limit load are: Male part: R.F. =

[

1

]

=

]

=

0,625 R 1a,6 + R 1tr,6

1

éæ êç Fa êç Pbry ëè

ö ÷ ÷ ø

1, 6

ö ÷ ÷ ø

1, 6

æF + ç tr ç Ptry è

ö ÷ ÷ ø

1, 6 ù

0,625

=

1,6 ù

0, 625

=

ú ú û

1

éæ 4667 ö1,6 æ 2000 ö1,6 ù êç ÷ +ç ÷ ú è 7900 ø úû êëè 12940 ø

0, 625

= 2,09

0,625

= 2,33

Female part: R.F. =

[

R 1a,6

1

0, 625 + R 1tr,6

1

éæ êç Fa êç Pbry ëè

æF + ç tr ç Ptry è

ö ÷ ÷ ø

ú ú û

1

éæ 4667 ö êç ÷ êëè 14380 ø

1,6

æ 2000 ö +ç ÷ è 8780 ø

1, 6 ù

ú úû

V1.7 8.3 PIN STRESSING Pin shear:

Here, the pin is subjected to double shear; the allowable force is therefore equal to: Psu =

π ⋅ d 2 ⋅ τadm π × 182 × 52 = = 26465 daN 2 2

The force applied to the pin is: F = Fa2 + Ftr2 = 70002 + 30002 = 7616 daN

Hence the R.F. of the pin subjected to shear is: R.F. =

Psu 26465 = F 7616

Þ


R.F. = 3,47

Numerical Example

page V1-7•8-5

SSM V1-7 • LUGS Pin bending At ultimate load: The allowable oblique force at ultimate load Padm is:

Padm =

F

[

]

0,625 R 1a,6 + R 1tr,6

=


0, 625

= 11017 daN

The peaking parameter is defined from the graph on figure V1-7 .7.2.1-2

ü 18 ù é ï ê20 − 2 ú ë û = 0,61ïï 18 ý Þ γ = 0,38 ï 11017 ï = = 0,77 ïþ 324 × 44

é Dù êa − 2 ú ë û = r= t2 Padm A br ⋅ σ R

The lever arm to be considered at ultimate load is therefore: t ö æt æ 10 18 ö b = γ ⋅ çç 1 + 2 ÷÷ + g = 0,38 × ç + ÷ + 0,6 = 4,21 mm 2 4 è 2 4ø ø è For a solid circular pin, the Cozzone calculation gives: k=

16 ≈ 1,7 3π

We would like that at ultimate load the stress at the extreme fibre of the pin does not exceed the ultimate strength, that is: σ M = σR = 90 daN / mm2

We deduce the bending modulus σ0 : σ0 æ n − 1 ö æç ε p MAX ö÷ æç æ n − 1 ö æç ε p MAX ö÷ ö÷ = 2ç ⋅ 1− ç ÷⋅ ÷⋅ σ MAX è n + 2 ø çè ε MAX ÷ø çè è 2n + 1 ø çè ε MAX ÷ø ÷ø

Where:

ε p MAX =

1 n

and

ε MAX =

σR 1 90 1 + = + = 0,033 E n 11030 40

σ0 = 80.3 daN / mm²

The allowable force at ultimate bending load is therefore: π ×18 4 4 4Ι [σ R + σ 0 ⋅ (k − 1)] = 64 × [90 + 80,3 × (1,7 − 1)] Pfu = 4,21× 18 b⋅d

Pfu = 39769 daN The corresponding bending R.F. is therefore: P 39769 Þ R.F. = 5,22 R.F. = fu = F 7616 page V1-7•8-6 © Copyright AEROSPATIALE 1998

Numerical Example

Issue 1

SSM V1-7 • LUGS At limit load:

The allowable oblique force at limit load Padm is: Padm =

F

[

]

0 ,625 R 1a,6 + R 1tr,6

=

7616 / 1,5

éæ 4667 ö1,6 æ 2000 ö1,6 ù êç ÷ +ç ÷ ú è 7900 ø ûú ëêè 12940 ø

0 ,625

= 10630 daN

The peaking parameter is defined from the graph on figure V1-7 .7.2.1-2

ü 18 ù é ï ê20 − 2 ú ë û = 0,61ïï 18 ý Þ γ = 0,53 ï 10630 = = 1,08 ï 324 × 30,5 þï

é Dù êa − 2 ú ë û r= = t2 Padm A br ⋅ σ R

The lever arm to be considered at ultimate load is therefore: t ö æt æ 10 18 ö b = γ ⋅ ç 1 + 2 ÷ + g = 0,53 × ç + ÷ + 0,6 = 5,64 mm 4ø è2 4ø è 2

The allowable force at ultimate bending load is therefore:

Pfy =

4Ι ⋅ σ0, 2 b⋅d

=

4

π × 184 × 87 64 5,64 × 18

Pfy = 17664 daN The corresponding bending R.F. is therefore: R.F. =

Pfy F

=

17664 7616 / 1,5


Þ R.F. = 3,48

Numerical Example

page V1-7•8-7


YES

X

NO


1 4

5 4

2 1 2

Volume 1

5


Purpose

Scope


Contents

Document Manager



Not applicable.

V1 - 1 V1 - 2 V1 - 3 V1 - 4 V1 - 5 V1 - 6 V1 - 7 V1 - 8 V1 - 9



Validation

Name: J. HUET



3page 1

Title - Annex

Reference documents


Abbreviations




Definitions


Highlights

Issue

Date

Pages modified

A

02/98

V1 - 1 à V1 - 3 V1 - 7 à V1 - 9

B

05/99

V1 - 7


V1 - 4

New chapter.

V1 - 1


C

11/99




New chapter.

MTS 004 Iss. C

3Ann. page



List of approval

Dept. code

Function

BTE/CC/CM

Name / First name


Key words

Calcul

Bibliography

Néant

Signature

CAZET G.

Distribution list

Dept. code

Function


BQP/TE

Archives Diderot

SIBADE Alain

BQP/TE


SIBADE Alain


BOUTET Fernand

BTE/SM/MG



MTS 004 Iss. C

page IG1


V1-8 •

HOLE REINFORCEMENT

A - CONTENTS Chapters V1-8 1 V1-8 2 V1-8 3 V1-8 4 V1-8 5 V1-8 6

Titles Issues General 1 Data preparation 1 General method 1 Load in the fasteners of a fastened reinforcement 2 Block diagram 1 Examples 1

Dates 04/1997 04/1997 04/1997 01/1998 04/1997 04/1997

Original issue Original issue Original issue Modification Original issue Original issue Pages V1.8 / 1 V1.8 / 1 V1.8 / 2 V1.8 / 3

V1-8. HOLE REINFORCEMENT A - CONTENTS B - SYMBOLS C - REFERENCES

V1-8 • 1. GENERAL V1.8.1/1 V1-8 • 1.1. DEFINITIONS V1.8.1/1 V1-8 • 1.2. BEHAVIOUR AND FAILURE MODE V1.8.1/2 V1-8 • 1.3. PRESENTATION OF THE METHOD V1.8.1/2 V1-8 • 2. DATA PREPARATION V1.8.2/1 V1-8 • 2.1. MATERIAL DATA V1.8.2/1 V1-8 • 2.2. GEOMETRICAL DATA V1.8.2/1 V1-8 • 3. GENERAL METHOD V1.8.3/1 V1.8.3/1 V1-8 • 3.1. CALCULATION OF OVER-STRESS COEFFICIENT Kt V1.8.3/2 V1-8 • 3.1.1. Calculation of η for a compact reinforcement V1.8.3/2 V1-8 • 3.1.2. Calculation of η for a flanged edge V1.8.3/3 V1-8 • 3.1.3. Calculation of η for a wide reinforcement V1-8 • 3.1.4. Taking infinite stress state into account V1.8.3/4 V1-8 • 3.1.5. Conclusion V1.8.3/5 V1.8.3/6 V1-8 • 3.2. CALCULATION OF EQUIVALENT STRESS σequi V1-8 • 3.3. TAKING BLEND RADII INTO ACCOUNT V1.8.3/7 V1-8 • 3.4. LOAD IN THE FASTENERS OF A FASTENED REINFORCEMENT V1.8.3/8 V1-8 • 3.4.1. Circular hole V1.8.3/8 V1-8 • 3.4.2. Elliptical hole V1.8.3/10 V1-8 • 3.4.3. Rectangular hole V1.8.3/11 V1-8 • 4. BASIC DATA V1.8.4/1 V1.8.4/1 V1-8 • 4.1. CALCULATION OF η V1.8.4/8 V1-8 • 4.2. CALCULATION OF Kt V1-8 • 4.2.1. Wide reinforcement: circular hole V1.8.4/9 V1-8 • 4.2.2. Narrow reinforcement: elliptical and circular hole V1.8.4/15 V1-8 • 4.2.3. Narrow reinforcement: square hole V1.8.4/18 V1-8 • 4.2.4. Narrow reinforcement: triangular hole V1.8.4/21 V1-8 • 4.2.5. Narrow reinforcement: circular hole close to an edge V1.8.4/30 V1-8 • 5. BLOCK DIAGRAM V1.8.5/1 V1-8 • 6. EXAMPLES V1.8.6/1 V1-8 • 6.1. CIRCULAR HOLE WITH A WIDE SYMMETRICAL REINFORCEMENT V1.8.6/1 V1-8 • 6.2. ELLIPTICAL HOLE WITH A COMPACT NON-SYMMETRICAL NARROW REINFORCEMENT V1.8.6/3 V1-8 • 6.3. BUSH INSERTED AS A REINFORCEMENT INSIDE A CIRCULAR HOLEV1.8.6/4 V1-8 • 6.4. REINFORCEMENT FOR A TRIANGULAR HOLE V1.8.6/5 V1-8 • 6.5. EVOLUTIVE THICKNESS REINFORCEMENT V1.8.6/7




V1-8 •

HOLE REINFORCEMENT

B - SYMBOLS Kt max: d: D: tr: hr: t: br, ar: ra: rr: c: Kt: σmax: σequi: A: Ar: Ap: η: I∞: Ixx: Ep: Er: σx, σy, τ: σ1, σ2: Ktr: Ktb:

allowable theoretical coefficient circular hole diameter circular reinforcement diameter reinforcement width reinforcement height plate thickness dimensions defining elliptical, square, triangular holes corner radius for triangular holes corner radius for square holes edge-distance of a hole over-stress coefficient maximum equivalent stress gross equivalent stress (Von Mises) equivalent section radial section of the reinforcement radial section of the plate under the reinforcement geometrical efficiency coefficient of the reinforcement inertia of the reinforcement in relation to a centreline passing through its centre of gravity and parallel to the neutral line of the plate inertia of the reinforcement in relation to the neutral line of the plate Young's modulus of the plate Young's modulus of the reinforcement stress condition around the hole main stresses over-stress coefficient at the joint of the reinforcement and the plate for a circular hole over-stress coefficient at the hole edge for a circular hole




V1-8 •

HOLE REINFORCEMENT

C - REFERENCES

REF. 1:

Elastic stress concentration factors. Single reinforced and unreinforced holes in infinite plate of isotropic materials ESDU DATA ITEM 80027

REF. 2:

Stress concentration factors R.E. PETERSON




V1-8 •

HOLE REINFORCEMENT





V1-8 •

HOLE REINFORCEMENT

V1-8 • 1.GENERAL This document proposes a method to calculate the maximum stress around a reinforced hole in an infinite plate or close to a free edge. The plate is assumed stable.

V1-8 • 1.1.DEFINITIONS Holes are frequently used in aircraft structures. They are used to lighten the structures, route cables or ducts or even to give access to certain other areas of the structure. Generally, these holes are reinforced by extra thickness, a fastened reinforcement or an internal bush: - the extra thickness is machined in the plate with the hole, - the fastened reinforcement is a ring secured around the hole by bolts, rivets or tack welds fastened, - the bush in the hole is made of a material different from the plate. The advantage of this technique is to reinforce the hole without using an extra thickness (more compact). An extra thickness or a fastened reinforcement may (or may not) be symmetrical in relation to the neutral line of the plate.

V1-8 • 1.2.BEHAVIOUR AND FAILURE MODE A hole in a structure locally modifies the mechanical behaviour of the structure. In particular, the presence of a hole creates an over-stress state around it. This over-stress may exceed the allowable limit of the material and cause local failure in the structure. Use of a hole reinforcement is recommended in this case. The stress state around the reinforced hole is taken into account using the Mises-Hencky criterion. The method is used to calculate the maximum equivalent stress that can be compared with the selected allowable limit. Note: In regular structural areas, a theoretical over-stress coefficient is accepted (in relation to the gross stress): Kt max = 2.4




V1-8 •

HOLE REINFORCEMENT

V1-8 • 1.3. PRESENTATION OF THE METHOD To cover as many cases as possible, this study will take different geometries of holes and of reinforcements under different stress states into account. The various hole geometries are: - circular holes - elliptical holes - rectangular holes (with rounded corners) - triangular holes (with rounded corners) The different types of hole reinforcements are: - extra thickness - fastened reinforcement (for example riveted) - bush inside the hole (bush and plate materials different) They are defined using two geometrical criteria: - D/d: the ratio between the hole diameter and the diameter of the reinforcement, - hr/tr: the ratio between the reinforcement height and its width. t∞ h∞

d D

If D/d > 1.05: the reinforcement is considered to be sufficiently "wide" to take the variation in the stress along a radial cross section of the reinforcement into account. In this case, the Kt max at the edge of the hole and at the joint between the reinforcement and the plate is calculated. If D/d < 1.05: the reinforcement is considered "narrow". This means that the stress is assumed to be uniform along a radial cross section. The reinforcement will be represented by its surface area and the interface with the plate by a line. This line will be defined by the outer radius of the reinforcement. The Kt max will be calculated at some point of this line. In this case, two sub-types of narrow reinforcements are differentiated: If hr/tr < 3: the reinforcement is considered as being compact. If hr/tr > 3: the reinforcement is considered as being a flanged edge.

WIDE D/d > 1.05

DIFFERENT TYPES OF REINFORCEMENTS NARROW D/d < 1.05 COMPACT: hr/tr < 3 FLANGED EDGE: hr/tr > 3

D

d

D hr

d tr


D hr

d tr



V1-8 •

HOLE REINFORCEMENT

In general, a hole reinforcement is a local over-thickness around the hole. For reasons of overall dimensions another reinforcement principle may be used. This principle consists in inserting a bush, made of a material different from the plate, in the hole. The method described in this chapter covers this type of reinforcement as it takes the differences in Young's modulus and Poisson ratio between the plate and the reinforcement into account. Also, the fastened reinforcement case (for example riveted) is handled as an extrapolation of the general method discussed in this chapter. The different load cases are taken into account by means of the Mises-Hencky criterion. The method is used to calculate the maximum equivalent stress that can be compared with the selected allowable limit. The elementary loads studied here are: - uniaxial tension, - biaxial tension, - shear. The calculated stress for a combined load will be (by interpolation) a linear combination of stresses defined for each basic case. Remarks: • In the case of a circular hole, it if preferable to use the case of the main stresses in biaxial tension (refer to paragraph V1-8.3.1.4). • The real stress is always less than or equal to this calculated stress. In fact, the place where the maximum stress is located is not the same for each load case. The combination of these maximum stresses is therefore greater than the real stress.




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HOLE REINFORCEMENT

V1-8 • 2.DATA PREPARATION V1-8 • 2.2.MATERIAL DATA The Young's modulus:E must be known for each of the materials used (plate and reinforcement).

V1-8 • 2.2.GEOMETRICAL DATA The geometrical data required for the study is specific to each type of reinforcement and hole discussed in this chapter. • For a wide reinforcement and a circular hole, the following must be known: t t+tr

dD

• For a narrow reinforcement and an elliptical or circular hole, the following must be known: t br tr ar

Data preparation - Page V1-8•2/1

hr

revision 1 © Copyright AEROSPATIALE 1998


V1-8 •

HOLE REINFORCEMENT

• For a narrow reinforcement and a square hole, the following must be known: t 2br tr rr hr

• For a narrow reinforcement and a rectangular hole, the following must be known: t tr

ra hr

br

• For a narrow reinforcement and a circular hole close to an edge, the following must be known: t tr ar

hr

c

Data preparation - Page V1-8•2/2



V1-8 •

HOLE REINFORCEMENT

V1-8 • 3.GENERAL METHOD A hole in an infinite plate. As this hole is reinforced (over-thickness, bush, added ring) the goal is to calculate the maximum stress in the hole reinforcement area knowing the stress field in the plate close to the hole. To this end, the over-stress coefficient Kt and Mises-Hencky criterion are used to calculate the equivalent stresses. The maximum equivalent stress due to the presence of the reinforced hole is: σmax = Ktσequi with: σmax: maximum equivalent stress (to be compared with the limit stress) Kt: over-stress coefficient σequi: gross equivalent stress (Von Mises) around the reinforcement area (function of the stress state in the plate) Therefore, it is necessary to calculate the over-stress coefficient Kt and the equivalent stress σequi as a function of the geometry and the stress state in the plate.

V1-8 • 3.1.CALCULATION OF OVER-STRESS COEFFICIENT Kt This coefficient depends on the geometry of the hole, the type of reinforcement and the stress state in the plate around the hole. To characterize the geometry of the hole and the type of reinforcement, the equivalent section of the reinforcement is defined. To solve the real case, this parameter, noted A , makes it possible to use the results obtained in basic cases. The formulation of this equivalent section is: A = Ap + ηAr

with: Ap: Ar: η:

cross section area of the plate under the reinforcement reinforcement cross section area geometrical efficiency coefficient of the reinforcement

The calculation of η differs according to the type of reinforcement. This coefficient takes both the neutral fibre offset (if the reinforcement is not symmetrical in relation to the plate) and the differences in the materials of the plate and the reinforcement (if this is the case) into account.

General method - Page V1-8•3/1



V1-8 •

HOLE REINFORCEMENT

V1-8 • 3.1.1.Calculation of η for a compact reinforcement The formulation of η for a compact reinforcement (D/d<1.05 and hr/tr<3) is: I E η = 00 r I xx E p with: I00: inertia of the reinforcement in relation to an axis parallel to the neutral line of the plate and passing through its centre of gravity Ixx: inertia of the reinforcement in relation to the neutral line of the plate Ep: Young's modulus of the plate Er: Young's modulus of the reinforcement

Example: a non-symmetrical rectangular cross section reinforcement: tr hr d/2

tp

D/2

In this case there is: t r h 3r I00 = 12

t r h 3r æ hr + tp ö + trhr and Ixx = I00 + Ar ç ÷ = è 2 ø 12 2

being: η =

1 tp ö æ 1 + 3 ç1 + ÷ hr ø è

2

æ hr + tp ö ç ÷ è 2 ø

2

Er Ep

If in addition the thickness of this plate in relation to that of the reinforcement is not taken into account, then η = 1/4

V1-8 • 3.1.2.Calculation of η for a flanged edge The charts in chapter V1-8.4.1 give the values of η for symmetrical and non-symmetrical flanged edges in biaxial tension. The thin plate theory has been used to plot these curves and the regions in which this theory is not truly applicable are shown in dotted lines. These η values may be used to calculate non-circular holes with any stress state. The generated errors are low if the point where the maximum stress is located is on a rounded contour and if the hole is practically circular.




V1-8 •

HOLE REINFORCEMENT

Remark: The cross section area of the plate under the reinforcement, Ap, participates in the reinforcement. If the Young's modulus of the reinforcement and that on the plate are different, then this has to be taken into account in the calculation of η. To this end, it is necessary to multiply the value read on the chart by the ratio: Er/Ep. If the plate and the reinforcement are made of different materials, then: ηtrue = ηchart

Er Ep

with: Ep: Er:

Young's modulus of the plate Young's modulus of the reinforcement

The cases shown in the charts in Chapter 1.8.4.1 are: - symmetrical flanged edge, - non-symmetrical flanged edge for hr/tr values between 3 and 10.

V1-8 • 3.1.3.Calculation of η for a wide reinforcement The results obtained on the large reinforcements proposed in this chapter, do not make it possible to take a non-symmetrical reinforcement or a difference in material between the plate and the reinforcement into account. In the case of a large non-symmetrical reinforcement or if the material is different from that of the plate, two approximations are possible: - the reinforcement may be considered as being narrow. The advantage is to be able to take into account the non-symmetry and the difference in material by calculating η (refer to paragraph V1-8.3.1.1). The disadvantage is having to calculate the stress at joint between the reinforcement and the plate assuming it to be constant throughout the reinforcement. - the reinforcement may be considered as being wide, symmetrical and of the same material as the plate (refer to the example, paragraph V1-8.6.5). The advantage is that it takes the development of the stress in the reinforcement into account by calculating it at the edge of the hole and at the joint of the reinforcement and the plate. The disadvantage is that the neutral fibre offset or the difference in material cannot be taken into account. These approximations are also necessary for a non-circular wide reinforcement. In fact, only the case of a wide circular reinforcement is discussed in this chapter.




V1-8 •

HOLE REINFORCEMENT

V1-8 • 3.1.4.Taking infinite stress state into account For the calculation of Kt, using the charts in Chapter 1.8.4.2, in addition to the calculation of the equivalent cross section area A , it is also necessary to take into account the gross stress state around the reinforcement. For wide reinforcements (D/d>1.05), it is necessary to calculate the main gross stresses (in the plate, without a hole and without reinforcement):

τ

τ

σy

τ

σ2

τ

σ1

σx

σx σ1

σ2

τ

τ τ

There is:

τ

σy

[

]

1 2 1 σx − σy ) + τ2 (σx + σy) + ( 4 2 2 1 1 σ2 = (σx + σy) σx + σy + τ2 4 2 σ1 =

[(

)

]

It is then necessary to obtain the linear combination which makes it possible to calculate these main stresses using the processed basic cases. Example: For the case of a circular hole reinforced by a wide ring (D/d>1.05), five basic cases are proposed: σ2 σ1 case 1 1 1 case 2 1 0.5 case 3 1 0 case 4 1 -0.5 case 5 1 -1 σ 1 If for the case to be processed, there is: 2 = σ1 4 K t ( cas3 ) + K t ( cas2 ) then: K æ σ 1 ö = with the Kt(cas i) obtained from the charts. 2 tç 2 = ÷ 4 σ è 1

ø




V1-8 •

HOLE REINFORCEMENT

For narrow reinforcements (D/d<1.05), unlike wide reinforcements, it is not necessary to calculate the main stresses. All that is necessary is to obtain the linear combination of the basic cases making it possible to calculate the real case.

V1-8 • 3.1.5.Conclusion Knowing the equivalent cross section area A , and the linear combination between the basic cases, making it possible to obtain the real stress status, it is possible to use the charts in Chapter 1.8.4.3.2 to determine Kt max.




V1-8 •

HOLE REINFORCEMENT

V1-8 • 3.2.CALCULATION OF EQUIVALENT STRESS σequi Let us assume any stress state around the hole.

τ

σy

τ

τ

τ

σx

σx

τ

τ τ

σy

τ

Then: the formulation of the equivalent stress: (Mises-Hencky criterion) is: σequi = σ 2x − σ x σ y + σ 2y + 3τ 2

This equivalent stress multiplied by the coefficient Kt makes it possible to calculate the maximum stress at the edge of the hole or around the reinforcement. This maximum value can be directly compared to the fixed limit value.

The basic data required to calculate Kt is given in the following chapter.




V1-8 •

HOLE REINFORCEMENT

V1-8 • 3.3.TAKING BLEND RADII INTO ACCOUNT With machined reinforcements, the presence of blend radii reduces the over-stress at the joint between the reinforcement and the plate. The following rule may be used to assess this effect. 1 - Calculation of Kt0 at the reinforcement/plate joint, in the case of a reinforcement without a blend radius, using the methods and charts given in this chapter. 2 - Taking the effect of the radius on Kt into account, using the following formula: Kt =β Kt0

with: - Kt0: over-stress coefficient calculated at the joint between the reinforcement with the plate without taking the presence of the radius into account. - β: correction coefficient depending on the geometry of the blend radius. This coefficient is calculated using the following chart: 1 0.95 0.9 0.85 0.8 0.75 β 0.7 0.65 0.6 0.55 0.5 0.45 0.4 0

0.25

0.5

0.75

1

1.25

1.5

1.75

2

r t

r


t



V1-8 •

HOLE REINFORCEMENT

V1-8 • 3.4.LOAD IN THE FASTENERS OF A FASTENED REINFORCEMENT The determination of the loads transferred by the fasteners of a fastened reinforcement is a relatively complex problem. Numerous parameters are to be taken into account such as the geometry of the hole, the type of reinforcement, the quantity, the location and the type of fasteners, the state of stress around the reinforcement. The method proposed here is an approximate method. It is simple but conservative. This means that it may induce very significant loads in fasteners. It is therefore preferable to use a local finite element model to get closer to reality. Assumptions: A hole reinforced in an infinite plate. The stress state around this reinforcement is any value. The different hole geometries are those proposed in this chapter (circular, elliptical, rectangular). Method: The main difficulty is to take shear stresses into account. An adapted method is proposed for each type of geometry. These methods make it possible to take each type of stress (normal or shear) into account separately.

V1-8 • 3.4.1.Circular hole σy

τ

t tr

σ2 σ1 σx

σx

d

D

σ2 σ1 τ σy The method is as follows: 1 - Calculation of main stresses as a function of infinite gross stresses: 2 1 1 σx − σy + τ2 σ1 = (σx + σy) + 4 2 2 1 1 σ2 = (σx + σy) σx − σy + τ2 2 4

[( [(


] )] )



V1-8 •

HOLE REINFORCEMENT

2 - Calculation of the total gross load transferred at the reinforcement in the main directions: Ftot1 = σ1 (D . t) Ftot2 = σ2 (D . t) 3 - Calculation of the loads transferred into the reinforcement in proportion to the thicknesses: æ t ö Frenf1 = Ftot1 ç r ÷ è t + tr ø

æ t ö Frenf2 = Ftot2 ç r ÷ è t + tr ø 4 - Distribution of each of these loads in each fastener. Distribution will occur at the same bearing and same shear as indicated in Chapter V1-6 "Splices" in this manual. For each main direction, a loading area and an unloading area will be considered. Fasteners subjected to F renf 1

σy

τ

Fasteners subjected to F renf 2

τ

σ2

τ

σ2 σ1

σx

σx

σ1

σ2

σ1 τ

σy

σx

σx

σ2

σ1 τ

σy

τ

τ

σy

τ

In this manner, two loads per fastener F/fix1 and F/fix2 are obtained which then simply have to be combined to obtain the total load per fastener: F/fix = F/2fix1 + F/2fix 2




V1-8 •

HOLE REINFORCEMENT

V1-8 • 3.4.2.Elliptical hole σy

τ

t

br tr ar

σx

τ

σx

hr

σy

The main stresses cannot be used for an elliptical hole. The normal stresses (σx, σy), are taken into account in the same manner as in the case of a circular hole with main stresses (refer to paragraph V1-8.3.4.1). The shear stresses (τ) are taken into account using the following method: 1 - Calculation of the total gross load related to the shear at the reinforcement: τ

Fτ = τ (L . t)

L

τ

2 - Distribution of this load between the plate and reinforcement in proportion to the thicknesses: æ hr ö Fτ renf = Fτ ç ÷ è t + hr ø 3 - Distribution of Fτ renf (force transferred into the reinforcement). On each fastener, distribution will occur at the same bearing and the same shear as indicated in Chapter V1-6 "Splices" in this manual. A loading area and an unloading area will be considered. The unloading area will be distributed to the following fasteners: τ

L

fastener subjected to Fτ renf

τ




V1-8 •

HOLE REINFORCEMENT

V1-8 • 3.4.3.Rectangular hole σy

τ

t

2br tr σx

σx

hr

2ar

τ

σy

Main stresses cannot be used for a rectangular hole. Normal stresses (σx, σy) are taken into account using the following method: 1 - Calculation of the total gross load transferred at the reinforcement in σx and σy directions. Ftotx = σx (2ar . t) Ftoty = σy (2br . t) 2 - Calculation of the loads transferred into the reinforcement in proportion to the thicknesses: æ hr ö Frenfx = Ftot1 ç ÷ è t + hr ø

æ hr ö Frenfy = Ftot2 ç ÷ è t + hr ø 3 - Distribution of each of these loads. Distribution will occur at the same bearing and the same shear as indicated in Chapter V1-6 "Splices" in this manual. For each main direction, a loading area and an unloading area will be taken into account: σy

fasteners subjected to Frenfy

σx

σx

fasteners subjected to Frenfx

σy

Note: Unlike the other fasteners, corner fasteners are subjected to both rFenfx and Frenfy.




V1-8 •

HOLE REINFORCEMENT

Shear stresses (τ) are taken into account using the following method: 1 - Calculation of gross loads related to shear at the reinforcement on each side: 2br

τ

Fτa = τ(2ar . t) Fτb = τ(2br . t)

2ar

τ

2 - Distribution of these loads between the plate and the reinforcement in proportion of the thicknesses: æ hr ö Fτ renf a = Fτa ç ÷ è t + hr ø

æ hr ö Fτ renf b = Fτb ç ÷ è t + hr ø and Fτ renf b (forces transferred to the reinforcement) on each

3 - Distribution of Fτ renf a fastener. This distribution will occur at the same bearing and the same shear as indicated in Chapter V1-6 "Splices" in this manual. The loads are assumed to be distributed on the following fasteners: 2br

fasteners subjected to F τ renf b τ

fasteners subjected to F τ renf a 2ar

fasteners subjected to -Fτ renf a

τ

fasteners subjected to -Fτ renf b

Load combinations: Finally, for each fastener, it is necessary to combine the loads due to the various stresses (σx, σy, τ).




V1-8 •

HOLE REINFORCEMENT

V1-8 • 4.BASIC DATA V1-8 • 4.1.CALCULATION OF η

Basic data - Page V1-8•4/1


HOLE REINFORCEMENT

hr

V1-8 •

rr

tr

hr


Figure 1-8.4.1.1


Symmetrical flanged edge


V1-8 •

HOLE REINFORCEMENT

rr

tr

hr

t


Figure 1-8.4.1.2


Non-symmetrical flanged edge

hr =3 tr


V1-8 •

HOLE REINFORCEMENT

rr

tr

hr

t


Figure 1-8.4.1.3



hr =4 tr


V1-8 •

HOLE REINFORCEMENT

rr

tr

hr

t


Figure 1-8.4.1.4



hr =5 tr


V1-8 •

HOLE REINFORCEMENT

rr

tr

hr

t


Figure 1-8.4.1.5



hr =7 tr


V1-8 •

HOLE REINFORCEMENT

rr

tr

hr

t


Figure 1-8.4.1.6



hr = 10 tr



V1-8 •

HOLE REINFORCEMENT

V1-8 • 4.2.CALCULATION OF Kt The cases dealt with here enable calculation of Kt in all other load or geometry cases by applying the method described in the previous chapter (paragraph 1.8.3). These basic cases are as follows: TYPE OF REINFORCEMENT

TYPE OF HOLE

σy

τ D/d > 1.05

WIDE

τ CIRCULAR

TYPE OF LOADING

σ2

σ1

σ1 σx

σx σ1

τ

σ2

τ

σy τ σy

τ ELLIPTICAL

σx σx τ

τ

σx σx

D/d < 1.05

NARROW

τ τ

σy τ σy

τ TRIANGULAR (equilateral rounded corners)

σx σx τ σy τ

CIRCULAR (close to an edge)

σx


σ2

1

1

1

0.5

1

0

1

-0.5

1

-1

1-8.4.2.1.1 1-8.4.2.1.2 1-8.4.2.1.3 1-8.4.2.1.4 1-8.4.2.1.5

σx

σy

τ

1

0

0

1

1

0

1

0.5

0

0

0

1

σx

σy

τ

1

0

0

1

1

0

1

0.5

0

0

0

1

σx

σy

τ

1

0

0

0

1

0

1

1

0

1-8.4.2.2.1 1-8.4.2.2.1 1-8.4.2.2.2 1-8.4.2.2.2

σy τ σy

τ SQUARE (rounded corners)

CHART NUMBERS

σx

1

0.5

0

0.5

1

0

0

0

1

σx only

1-8.4.2.3.1 1-8.4.2.3.1 1-8.4.2.3.2 1-8.4.2.3.2

1-8.4.2.4.2 1-8.4.2.4.3 1-8.4.2.4.4 1-8.4.2.4.5 1-8.4.2.4.6 1-8.4.2.4.7

1-8.4.2.5.1



V1-8 •

HOLE REINFORCEMENT

V1-8 • 4.2.1.Wide reinforcement: circular hole a) Necessary geometrical data: σy τ

t t+tr

σ2

σ1 σx

σx

d

D

σ2 σ1 τ

σy

b) Use of the charts: The following charts enable the calculation of the over-stress coefficient at the edge of a hole (KtA) and at the joint of the reinforcement and the plate (K tB). These coefficients are given as a tr D function of the following ratios: and d t Remarks: D - if < 1.05 then refer to paragraph V1-8.4.2.2 d - if the reinforcement is not symmetrical, refer to paragraph V1-8.4.2.2 or consider that the radial cross section of the reinforcement is uniformly distributed on either side of the reinforcement by calculating an equivalent thickness rt (refer to the example in paragraph V1-8.6.5). - if the hole is elliptical, refer to paragraph V1-8.4.2.2 or consider an equivalent circular hole with an average radius d (refer to the example in paragraph V1-8.6.5). c) Single loads: The charts in this paragraph correspond to the following single load cases. TYPE OF LOADING σ1 1 1 1 1 1

σ2 1 0.5 0 -0.5 -1

CHART NUMBERS

1-8.4.2.1.1 1-8.4.2.1.2 1-8.4.2.1.3 1-8.4.2.1.4 1-8.4.2.1.5

For any other type of load, it is possible to calculate the maximum stress by combining these basic cases (conservative calculation). This method is used in the example in paragraph V18.6.1.




V1-8 •

σy σ2

τ

σ1

τ

t t+tr

σ1

A

σx

HOLE REINFORCEMENT

σx

B

d

D

σ2 σy

Figure 1-8.4.2.1.1




V1-8 •

HOLE REINFORCEMENT σy σ2

τ σ1

A

σx σ1

τ

t t+tr

σx

B

d

D

σ2 σy

Figure 1-8.4.2.1.2




V1-8 •

HOLE REINFORCEMENT

t t+tr σ1

σ1 A

d

D

B

Figure 1-8.4.2.1.3




V1-8 •

HOLE REINFORCEMENT σy

σ2

τ σ1

A

σx σ1

τ

t t+tr

σx

B

d

D

σ2 σy

Figure 1-8.4.2.1.4




V1-8 •

HOLE REINFORCEMENT σy

σ2

τ σ1

A

σx σ1

τ

t t+tr

σx

B

d

D

σ2 σy

Figure 1-8.4.2.1.5




V1-8 •

HOLE REINFORCEMENT

V1-8 • 4.2.2.Narrow reinforcement: elliptical and circular hole a) Necessary geometrical data: σy

τ t

br tr ar

σx

σx

hr

τ σy

b) Use of the charts: The following charts are used to calculate the over-stress coefficient at the joint between the reinforcement and the plate (Kt). These coefficients are given as a function of the following a A and r ratios: br (a r + b r ) t Remarks: - A is the equivalent section of the reinforcement, which takes the non-symmetry of the reinforcement or a difference in material between the reinforcement and the plate into account where applicable. c) Single load case: The charts in this paragraph correspond to the following single load cases. TYPE OF LOADING σx 1 1 1 0

σy 0 1 0.5 0

CHART NUMBERS τ 0 0 0 1

1-8.4.2.2.1 1-8.4.2.2.1 1-8.4.2.2.2 1-8.4.2.2.2

For any other type of loading, it is possible to calculate the maximum stress by combining these basic cases (conservative calculation). This method is used in the example in paragraph V1-8.6.1.




V1-8 •

HOLE REINFORCEMENT

ar σx

σx

br σy = 0 τ=0

σy

ar σx

σx

br σy σy σx

=1

τ=0

Figure 1-8.4.2.2.1




V1-8 •

HOLE REINFORCEMENT

σy

ar σx

σx

br σy

σy σx

=

1 2

τ=0

τ ar

br

τ σx = 0 σy = 0

Figure 1-8.4.2.2.2




V1-8 •

HOLE REINFORCEMENT

V1-8 • 4.2.3.Narrow reinforcement: square hole a) Necessary geometrical data: σy

τ

t

2br

tr σx

rr

σx

τ

hr

σy

b) Use of the charts: The following charts are used to calculate the over-stress coefficient at the joint between the reinforcement and the plate (Kt). These coefficients are given as a function of the following r A ratios: and r br brt Remarks: - A is the equivalent section of the reinforcement, which takes the non-symmetry of the reinforcement or a difference in material between the reinforcement and the plate into account where applicable. c) Single load case: The charts in this paragraph correspond to the following single load cases. TYPE OF LOADING

σx 1 1 1 0

σy 0 1 0.5 0

CHART NUMBERS

τ 0 0 0 1

1-8.4.2.3.1 1-8.4.2.3.1 1-8.4.2.3.2 1-8.4.2.3.2





V1-8 •

HOLE REINFORCEMENT

2br rr

σx

σx

σy = 0 τ=0

σy 2br rr

σx

σx

σy σy

=1

σx

τ=0

Figure 1-8.4.2.3.1




V1-8 •

HOLE REINFORCEMENT

σy 2br rr

σx

σx

σy σy

=1

σx

2

τ=0

τ 2br rr

τ σx = 0 σy = 0

Figure 1-8.4.2.3.2




V1-8 •

HOLE REINFORCEMENT

V1-8 • 4.2.4.Narrow reinforcement: triangular hole a) Necessary geometrical data: σy τ

t tr

ra σx

σx

br

hr

τ σy

b) Use of the charts: The following charts are used to calculate the over-stress coefficient at the joint between the reinforcement and the plate (Kt). r r Initially, using the ratio a , it is necessary to determine the theoretical ratio t using the chart br br in Figure 1-8.4.2.4.1. r A Then the over-stress coefficients are given as a function of the ratios: and t . br brt Remarks: - A is the equivalent section of the reinforcement, which takes the non-symmetry of the reinforcement or a difference in material between the reinforcement and the plate into account where applicable. c) Single load case: The charts in this paragraph correspond to the following single load cases. TYPE OF LOADING No. 1 2 3 4 5 6

σx 1 0 1 1 0.5 0

σy 0 1 1 0.5 1 0

CHART NUMBERS τ 0 0 0 0 0 1

1-8-4.2.4.2 1-8.4.2.4.3 1-8.4.2.4.4 1-8.4.2.4.5 1-8.4.2.4.6 1-8.4.2.4.7





V1-8 •

HOLE REINFORCEMENT

d) Location of the over-stress:

B A B y x

The position of the over-stress and the joint of the reinforcement and the plate depends on the stress status around the reinforcement. For cases 1, 3 and 4, the over-stress is at corners B. For case 5, the over-stress is at corner A. For cases 2 and 6, this position depends on the geometrical ratios of the triangle:

ra rt , and br br

A . To locate this over-stress, regions have been defined on the corresponding charts: Figure brt 1-8.4.2.4.3 and Figure 1-8.4.2.4.7. Region 1 corresponds to an over-stress at corners B and region 2 corresponds to an over-stress at corner A. TYPE OF LOADING

LOCATION

No. 1 2

σx 1 0

σy 0 1

τ 0 0

3 4 5 6

1 1 0.5 0

1 0.5 1 0

0 0 0 1


B B (if region 1) A (if region 2) B B A B (if region 1) A (if region 2)



V1-8 •

HOLE REINFORCEMENT

ra br

real shape

rt br

theoretical shape

Figure 1-8.4.2.4.1




V1-8 •

ra

HOLE REINFORCEMENT

br

σy = 0

σx τ = 0

σx

Figure 1-8.4.2.4.2




V1-8 •

HOLE REINFORCEMENT

σy

ra

br

σx = 0 τ=0

σy

Figure 1-8.4.2.4.3




V1-8 •

HOLE REINFORCEMENT

σy ra

br σx

σx

σx = σy τ=0

σy

Figure 1-8.4.2.4.4




V1-8 •

HOLE REINFORCEMENT

σy ra

br

σy 1 σx

σx

σx 2 τ=0

σy

Figure 1-8.4.2.4.5




V1-8 •

HOLE REINFORCEMENT

σy ra

br

σx σx σ y

σx

=

1 2

τ=0 σy

Figure 1-8.4.2.4.6




V1-8 •

HOLE REINFORCEMENT

τ ra br

σx = 0 σy = 0

τ

Figure 1-8.4.2.4.7




V1-8 •

HOLE REINFORCEMENT

V1-8 • 4.2.5.Narrow reinforcement: circular hole close to an edge a) Necessary geometrical data:

t tr hr

ar

c

b) Use of the charts:

A c , the following chart is used to calculate the over-stress and ar 2a r t coefficient at the joint of the reinforcement and the plate (K tr) and at the edge of the plate (Ktb). Depending on the ratios:

Remarks: - A is the equivalent section of the reinforcement, which takes the non-symmetry reinforcement or a difference in material between the reinforcement and the plate into account where applicable. c) Single load case: The chart in this paragraph corresponds to the unidirectional load case. TYPE OF LOADING σx 1

σy 0


CHART NUMBERS τ 0

1-8.4.2.5.1



V1-8 •

ar

σx

HOLE REINFORCEMENT

σx

c

Figure 1-8.4.2.5.1




V1-8 •

HOLE REINFORCEMENT





V1-8 •

HOLE REINFORCEMENT

V1-8 • 5.BLOCK DIAGRAM CALCULATION OF MAXIMUM STRESS AT THE EDGE OF THE REINFORCED HOLE

CALCULATION OF σequi

CALCULATION OF Kt maxi

D/d > 1.05

yes

no Calculation of η

hr/tr > 3

yes ηreal = ηchart

Er Ep

no

I 00 E r I xx E p

η=

Calculation of σ1 and σ2

[( σ 4

1 σ1 = (σx + σy) + 2

1

1 σ2 = (σx + σy) 2

1 4

[( σ

x

− σy )

x

− σy )

2

2

]+τ ]+τ

E S S E

2

Calculation of the equivalent cross section area A : A = Ap + ηAr

2

Kt(real) = combination (Kt(basic case)) Kt(basic case) using the charts

σequi = σ 2x − σ x σ y + σ 2y + 3τ 2

σmax = Ktσequi σy

τ τ σx

τ

τ

σ2

σ1

τ

σy

T I

τ σ1

σ2

N

A

σx τ

τ

Block diagram - Page V1-8•5/1 revision 1

L


V1-8 •

HOLE REINFORCEMENT


Block diagram - Page V1-8•5/2 revision 1


V1-8 •

HOLE REINFORCEMENT

V1-8 • 6.EXAMPLES V1-8 • 6.1.CIRCULAR HOLE WITH A WIDE SYMMETRICAL REINFORCEMENT Let us assume a 100 mm diameter hole in a 5 mm infinite plate. This hole is reinforced by a 200 mm diameter, 10 mm thick symmetrical over-thickness. σy

τ τ σx

σx

τ σy

τ

200 mm 10 mm 100 mm A

B

5 mm 10 mm

The stress state around the hole is as follows: σx = 185 Mpa σy = 65 Mpa τ = 45 Mpa D 200 = = 2 > 1.05, therefore it is a wide reinforcement. d 100 In this case, it is necessary to consider a development of the stress according to the reinforcement width (between A and B). To determine the over-stress at the edge of the hole (at A) and at the reinforcement/plate joint (at B), the charts in Chapter V1-8.4.2.1 will be used after calculating the main stresses around the hole.

There is:

Examples - Page V1-8•6/1



V1-8 •

HOLE REINFORCEMENT

• Calculation of main stresses and equivalent stress: 1 1 1 1 σ 1 = (σ x + σ y ) + σx − σy + τ2 σ 1 = (185 + 65) + [185 − 65] + 45 2 2 4 2 4 being 1 1 1 1 σ 2 = (σ x + σ y ) − σx − σy + τ2 σ 2 = (185 + 65) − [185 − 65] + 45 2 2 4 2 4 σ = 200 MPa Giving: 1 σ 2 = 50 MPa Similarly, the equivalent stress is: σequi = σ 12 − σ 1 σ 2 + σ 22 = 200 2 − 200 . 50 + 50 2 giving: σequi = 180 MPa

[ [

] ]

• Determination of the linear combination: σ 1 σ = 200 MPa There is: σ 1 = 50 MPa being: 2 = 2 σ1 4 This load case does not correspond to one of the charts on wide reinforcements. Therefore, the σ 1 linear combination of basic cases has to be found to obtain 2 = σ1 4 The basic cases processed are: σ1 1 1 1 1 1

case 1 case 2 case 3 case 4 case 5

Then, the linear combination is: K æ σ

1ö tç 2 = ÷ è σ1 4 ø

=

σ2 1 0.5 0 -0.5 -1

K t ( cas3 ) + K t ( cas2 ) 2

with the Kt(casi) found in the

charts. • Calculation of the over-stress coefficient: Using the charts in Chapter V1-8.3.2.1, the following is obtained (with D = 2): d K t B(cas2) K t B(cas3) K t A (cas2) - At the edge of the reinforcement (pt A): K t A (cas3)

- At the edge of the hole (pt B):

t r 2 . 10 = = 4 and 5 tp

= 1.06 Þ K = 1.06 + 1.22 Þ K = 114 . tB tB 2 = 1.22 = 1.28 1.28 + 1.63 Þ K t A = 1.45 = 1.63 Þ K t A = 2

• Calculation of the maximum stresses at the edge of the hole and at the edge of the reinforcement: Using the equivalent stress and over-stress coefficients, the maximum stress values at the edge of the hole and at the edge of the reinforcement can be deduced: σmax B = Kt B . σequi= 1.14 . 180Þ σmax B = 205 Mpa - At the edge of the hole (pt B): - At the edge of the reinforcement (pt A): σmax A = Kt A . σequi= 1.45 . 180Þ σmax A = 261 Mpa




V1-8 •

HOLE REINFORCEMENT

V1-8 • 6.2.ELLIPTICAL HOLE WITH A COMPACT NON-SYMMETRICAL NARROW REINFORCEMENT τ τ 160 80

τ

τ 3 9

3

with: τ = 70 MPa In this case there is: D 83 = = 1.037 < 1.05 80 d • and h consequently, this is a compact narrow reinforcement 6 r = =2<3 3 tr As this reinforcement is not symmetrical, it is necessary to take its geometrical effectiveness coefficient η into account when calculating it. • Calculation of coefficient η: • Inertia of the reinforcement in relation to the axis passing through its centre of gravity t h 3 3. 6 3 and parallel to the neutral line of the plate: I00 = r r = Þ I00 = 54 mm4 12 12 • Inertia of the reinforcement in relation to the neutral line of the plate: Ixx = I00 + Sr . d 2x = 54 + 18 . 4.5²Þ Ixx = 418.5 mm4 with: Sr: reinforcement surface area d x : distance of the reinforcement centre of gravity to the neutral line of the plate • Geometrical effectiveness of the reinforcement:




V1-8 •

HOLE REINFORCEMENT

I 00 54 = Þ η = 0.13 I xx 418.5 • Calculation of equivalent cross section area A :

η=

A = Ap + ηAr = (3 . 3) + 0.13 . (3 . 6)Þ A = 11.34

being:

A 11.34 = = 0.015 (a r + b r ) t p (160 + 80) . 3

This ratio makes it possible to use the chart in Chapter V1-8.4.2.2 concerning the calculation of the Kt coefficient at the edge of the reinforcement of an elliptical hole in the pure shear case. There is: Kt = 2.2 Let us assume an over-stress at the joint between the reinforcement and the plate equal to: σmax = Ktσequi = 2.2 . (70 3 ) Þ σmax = 266 MPa

V1-8 • 6.3.BUSH INSERTED AS A REINFORCEMENT INSIDE A CIRCULAR HOLE

σx

σx

D d

t

with: D = 160 mm d = 150 mm t = 5 mm σx = 120 Mpa The plate and the bush are made of different materials: Plate made of 7175 T7351: Ep = 71000 Mpa Bush made of 2024 T3511: Eb = 74500 Mpa




In this case there is:

V1-8 •

HOLE REINFORCEMENT

D 160 = 1.06 > 1.05. Consequently, this is a wide reinforcement. = d 150

Now, to take the difference in material between the plate and the bush into account, the reinforcement is considered as being narrow. The geometrical effectiveness coefficient η is calculated and the charts in paragraph V1-8.4.2.2 are used to calculate Kt. • Calculation of the geometry effectiveness coefficient η: I 00 E r E = r as in this case: I00 = Ixx. I xx E p E p 74500 = 1.05 Giving: η = 71000 There is η =

• Calculation of the equivalent cross section area A : A = Ap + η . Ar = 0 + 1.05 . (10 * 5)Þ A = 52.5

A 52.5 = 0.065 = (a r + b r ). t p (80 + 80) . 5 This ratio makes it possible to use the charts in Chapter V1-8.4.2.2 concerning the calculation of coefficient Kt at the edge of a narrow reinforcement in the case of a unidirectional stress state. There is: Kt = 2.2 (instead of 3 without reinforcement) Being an over-stress at the edge of a hole equal to: σmax = Ktσequi = 2.2 . 120 Þ σmax = 264 MPa being:

V1-8 • 6.4.REINFORCEMENT FOR A TRIANGULAR HOLE τ τ

30

80

τ τ

3 3


9



V1-8 •

HOLE REINFORCEMENT

As this reinforcement is not symmetrical, it is necessary to take its geometrical effectiveness coefficient η into account when calculating it. • Calculation of the coefficient η: • Inertia of the reinforcement in relation to the axis passing through its centre of gravity t r h 3r 3 . 6 3 and parallel to the neutral line of the plate: I00 = = Þ I00 = 54 mm4 12 12 • Inertia of the reinforcement in relation to the neutral line of the plate: Ixx = I00 + Sr . d 2x = 54 + 18 . 4.5²Þ Ixx = 418.5 mm4 with Sr: surface area of the reinforcement d x : distance of the reinforcement centre of gravity to the neutral line of the plate • Geometrical effectiveness of the reinforcement: I 54 η = 00 = Þ η = 0.13 I xx 418.5 • Calculation of the equivalent section A : A = Ap + ηAr = (3 . 3) + 0.13 . (3 . 6)Þ A = 11.34

• Calculation of the equivalent triangle: There is: ra = 15 mm² br = 15 mm²

therefore:

ra 30 = = 0.375 b r 80

With this ratio and the chart in paragraph V1-8.4.2.4, the following is found:

rt = 0.2. br

• Calculation of the over-stress at the edge of the reinforcement: To calculate Kt, the chart in paragraph V1-8.4.2.4 concerning the pure shear case is used, with: A 11.34 • = 0.05 = b r t 80 . 3 r • t = 0.2 br (instead of 4, without reinforcement) There is: Kt = 3




V1-8 •

HOLE REINFORCEMENT

V1-8 • 6.5.EVOLUTIVE THICKNESS REINFORCEMENT τ

τ

160

143

A

A

8

τ

τ

17

143

3 7 11 SECTION AA

There is: - D = 143 + 25 = 168 mm - d = 143 mm D 168 therefore: = 1.17 > 1.05, this is a wide reinforcement. However, it is not = d 143 symmetrical and therefore the problem is handled with a narrow reinforcement to take the offset of the neutral fibre between the plate and the reinforcement into account. • Calculation of coefficient η: 8 1

17 2

3

7

11

• Inertia of the reinforcement in relation to the axis passing through its centre of gravity and parallel to the neutral line of the plate: 8 . 83 I 001 = = 341.33 mm 4 ;S1 = 64 mm ² 4 12 Þ I00 = (341.33 + 64 . 1²) + (90.66 + 68 . 1²) = 563.9 mm 17 . 4 3 4 = 90.66 mm ;S 2 = 68 mm ² I 002 = 12




V1-8 •

HOLE REINFORCEMENT

• Inertia of the reinforcement in relation to the neutral line of the plate: Ixx = I00 + Sr . d 2x = 563.9 + 132 . 4.47²Þ Ixx = 3201 mm4 with Sr: reinforcement surface area d x : distance of the reinforcement centre of gravity to the neutral line of the plate • Geometrical effectiveness of the reinforcement: I 563.9 η = 00 = Þ η = 0.176 3201 I xx • Calculation of the equivalent cross section area A : A = Ap + η . Ar = (3 . 25) + 0.176 . (64 + 68)Þ A = 98.2 mm²

A 98.2 = 0.0927 = (a r + b r ). t p (185 + 168) . 3 This ratio makes it possible to use the chart in Chapter V1-8.4.2.2 concerning the calculation of coefficient Kt and the edge of a narrow reinforcement in the pure shear case. There is:

being:

Kt = 1.41 (instead of 2.2, without reinforcement) Being an over-stress at the edge of the hole equal to: σmax = Ktσequi = 1.41 . (70 3 ) Þ σmax = 171 MPa Considering that the reinforcement is narrow, only an over-stress on the joint between the reinforcement and the plate is obtained, and it is assumed that this stress is constant along the reinforcement width to the edge of the hole. To take a change to this stress into account, it is necessary to use the method described in the case of wide reinforcements. However, this method only applies to circular holes with symmetrical reinforcements. This is not the case here, so it is assumed that the surface area of the reinforcement is evenly distributed each side of the plate and that the elliptical hole is replaced by an equivalent round hole. τ

τ

τ

τ

160

??? 143

equivalent to A

A

A

8

A

τ

τ

17

τ

τ

17 t r'

143

3 7 11 SECTION AA


t' SECTION AA



V1-8 •

HOLE REINFORCEMENT

with: - t' so that the radial cross section area of the reinforcement is maintained: (2 . t' + 3) . 25 = 207 mm² 1 207 − 3ö÷ Þ t' = 2.64 mm Þ t' = æç è ø 2 25 a + b r 160 + 143 = 151.5 mm - r' such that: r' = r = 2 2

2 . t' 2 2. 64 . D 2 . (151.5 + 25) = = 1.76 and = 1.16 = t 3 d 2 . (151.5) Using these ratios and the chart in paragraph V1-8.4.2.1.5 on wide reinforcements in pure shear, the following is found: - at the edge of the reinforcement at the joint with the plate: K t = 1.46being σmax = 177 Mpa

Then there is

- at the edge of the hole:

Kt = 1.63being σmax = 198 Mpa

Remark: When the reinforcement is considered as being narrow and non-symmetrical, an over-stress is found at the edge of the reinforcement equivalent to that obtained, assuming that the reinforcement is wide and symmetrical. However, the wide reinforcement assumption makes it possible to take into account a change to the stress of the radial section of the reinforcement. Here, this phenomenon is not negligible as the stress at the edge of the hole is 10% greater than the stress at the edge of the reinforcement.




YES

X

NO


1 4

5 4

2 1 2

Volume 1

5


Purpose

Scope


Contents

Document Manager



Not applicable.

V1 - 1 V1 - 2 V1 - 3 V1 - 4 V1 - 5 V1 - 6 V1 - 7 V1 - 8 V1 - 9



Validation

Name: J. HUET



3page 1

Title - Annex

Reference documents


Abbreviations




Definitions


Highlights

Issue

Date

Pages modified

A

02/98

V1 - 1 à V1 - 3 V1 - 7 à V1 - 9

B

05/99

V1 - 7


V1 - 4

New chapter.

V1 - 1


C

11/99




New chapter.

MTS 004 Iss. C

3Ann. page



List of approval

Dept. code

Function

BTE/CC/CM

Name / First name


Key words

Calcul

Bibliography

Néant

Signature

CAZET G.

Distribution list

Dept. code

Function


BQP/TE

Archives Diderot

SIBADE Alain

BQP/TE


SIBADE Alain


BOUTET Fernand

BTE/SM/MG



MTS 004 Iss. C

page IG1


V1-9 •

CLEATS

CONTENTS Chapter V1-9 V1-9.1 V1-9.2 V1-9.3 V1-9.4 V1-9.5 V1-9.6 V1-9.7 V1-9.8

Title CLEATS GENERAL DATA PREPARATION CLEAT STIFFNESS CLEAT EFFECTIVENESS CHECKING PROCEDURE CALCULATION EXAMPLE Appendix 1: Determination of stiffness criterion Appendix 2: Discussion on stiffness (...)

Issue 1 1 1 1 1 1 1 1 1

Date 05 / 1997 05 / 1997 05 / 1997 05 / 1997 05 / 1997 05 / 1997 05 / 1997 05 / 1997 05 / 1997

Original issue Original issue Original issue Original issue Original issue Original issue Original issue Original issue Original issue Pages

V1-9 .1 GENERAL ............................................................................................... V1-9•1/1 V1-9 .2 DATA PREPARATION.......................................................................... V1-9•2/1 V1-9

.2.1 MATERIAL DATA................................................................................................ V1-9•2/1

V1-9

.2.2 GEOMETRICAL DATA ....................................................................................... V1-9•2/1

V1-9 3-CLEAT STIFFNESS................................................................................... V1-9•3/1 V1-9

.3.1 ASSUMPTIONS ..................................................................................................... V1-9•3/1

V1-9 .3.2 DISPLACEMENT AND STIFFNESS .................................................................. V1-9 .3.2.1 Cleat bending: Dlbending/Mz................................................................................. V1-9 .3.2.2 Cleat rotation: ∆lrotation/z .................................................................................... V1-9 .3.2.3 Cleat translation: ∆ltranslation/x ............................................................................. V1-9 .3.2.4 Bending of the flange bearing on the frame: ∆lbending ....................................... V1-9 .3.2.5 General equation of cleat stiffness: K .............................................................. V1-9 .3.2.6 Calculation example.........................................................................................

V1-9•3/1 V1-9•3/2 V1-9•3/4 V1/9•3/5 V1-9•3/5 V1-9•3/7 V1-9•3/8

V1-9 .4 CLEAT EFFECTIVENESS.................................................................... V1-9•4/1 V1-9

.4.1 STIFFNESS CRITERION..................................................................................... V1-9•4/1

V1-9

.4.2 NON-RIGID CLEATS ........................................................................................... V1-9•4/4

V1-9 .5 CHECKING PROCEDURE ................................................................... V1-9•5/1 V1-9 .6 CALCULATION EXAMPLE................................................................. V1-9•6/1

Page V1-9/1 Revision 1



V1-9 •

CLEATS





V1-9 •

CLEATS

SYMBOLS USED Notation cdg x, y, z F ll

(N) (mm)

L W ∆l ∆lbending/z ∆lrotation/z ∆ltranslation/x ∆lbending l2 l3 Es Isz

(mm) (mm) (mm) (mm) (mm) (mm) (mm) (mm) (mm) (MPa) (mm4)

Mf (Mfz(y)) σy v / v(y)

(N.mm) (MPa) (mm)

σ0

Unit

(MPa)

n

e ∆y Mf0 I0

(mm) (mm) (N.mm) (mm4)

v0

(mm)

El Ilz

(MPa) (mm4)

Description centre of gravity local axes associated with the cleat load applied to the cleat at the end fastener on the member distance between the end fastener in the frame and the fasteners in the stringer distance between each frame ("inter-frame pitch") distance between each stringer ("inter-stringer pitch") total deflection at the end of the cleat deflection due to cleat bending deflection due to cleat rotation deflection due to cleat translation deflection due to cleat flange bending width of the cleat flange in contact with the frame height of the cleat web (generally evolutive) elasticity modulus of the cleat material moment of inertia of the cleat around z in relation to the centre of gravity of its cross section bending moment (at abscissa "y" around z) normal stress in the cleat x dimension between the centre of gravity of the cleat cross section and the end fibre ① constant normal stress in the case of a beam of equal strength ② normal stress in the cleat at joint with the stringer ① number of segments to discretize the cleat in the bending calculation ② number of elastic supports on the frame flange between two rigid supports cleat thickness length of a discretization segment bending moment around z in the cleat at the root (y = 0) moment of inertia of the cleat around z in relation to the centre of gravity of its cross section at the root (y = 0) x dimension between the centre of gravity of the cleat cross section and the end fibre at the root (y = 0) elasticity modulus of the stringer material moment of inertia of the stringer around z in relation to the centre of gravity of its cross section



Static stressing manual Notation Sl θ l4 l5 Ibending G G0 m

Unit (mm²) (slope) (mm) (mm) (mm4)

P Rm γ

(N) (N)

Ecc Ity

(MPa) (mm4)

I' ty

(mm4)

Xm k kcr r Pcr K Kmini

(mm)

λ

(N) (N/mm) (N/mm)

V1-9 •

CLEATS

Description stringer cross section angle of rotation length of an equivalent beam for cleat flange bending calculation width of an equivalent beam for cleat flange bending calculation moment of inertia of the cleat flange for bending calculation centre of gravity of a cleat cross section centre of gravity of the cleat cross section at the root (y = 0) ① index from 1 to n for elastic support identification between two rigid supports ② number of spans between two supports axial compression load in the frame inner flange reaction at intermediate support "m" dimensionless parameter used for the calculation of the stiffness criterion elasticity modulus in compression of the frame material moment of inertia of the flange around y in relation to the centre of gravity of its cross section moment of inertia of the inner flange around y in relation to the centre of gravity of its cross section, taking the elastic foundation into account displacement of the member inner flange at an elastic support "m" parameter used in the stiffness criterion calculation critical value of k for P = Pcr integer critical buckling load for the member inner flange cleat stiffness minimum stiffness of the cleat for it to be considered as a rigid support in relation to member warping member inner flange end fixity factor




V1-9 •

CLEATS

REFERENCES 1. 2. 3. 4. 5.

Michael C.Y. Niu: Airframe Structural Design Roark's: Formulas for Stress & Strain Timoshenko: Théorie de la stabilité élastique (Elastic Stability Theory) Bruhn: Analysis and Design of Flight Vehicle Structure Vallat: Résistance des matériaux appliquée à l'aviation (Material Strength Applied to the Aircraft Industry)




V1-9 •

CLEATS




Static stressing manual V1-9

V1-9 •

CLEATS

.1 GENERAL INFORMATION

Cleats are elements of stiffened panels (refer to paragraph V1-1) with the following functions: - to join stiffeners and transverse members, - to prevent premature warping of members (refer to paragraph V1-3). With fuselage panels, cleats may also provide frame-skin joints. In the example below (floating frame fuselage panel) the cleats are in the form of angle sections, one of the flanges of which (evolutive section) is fastened to the stringer and the other to the frame. These joints are secured by several fasteners (see Figure V1-9 .1-1). Generally, these fasteners are rivets (stringer side) and bolts (frame side).

Stringer Cleat

Frame

W

y

L W

z

x

Figure V1-9 .1-1: Typical stringer/floating frame/cleat assembly The purpose of this chapter is to describe a cleat stiffness calculation and verification procedure to determine whether or not they can be assimilated to rigid supports with regard to member warping. This chapter is closely associated with the complete warping study discussed in Chapter V3-3. The definition of the cleat can be validated when the minimum stiffness criterion is satisfied. Before reaching this result, a series of iterations is necessary. The problem of not meeting the

General information

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CLEATS

stiffness criterion can be satisfactorily solved in two different and possibly complementary manners: 1) Redimension a stiffer cleat. 2) Accept that a cleat is flexible and recalculate the critical warping stress of members using a longer beam segment. However, these two solutions must not be used inconsiderably: - in the first case, it may be difficult to obtain the required stiffness, especially as stiffness does not depend on cleats alone, - in the second case, there is the risk of lowering the critical warping stress of the frame flange too much. Then, it may be possible to envisage reinforcing the inner flange of the members. The static strength of fuselage panel cleats is checked using the pressurization loads transferred from the fuselage to the frame. This is discussed in a separate chapter. The following mechanism must be fully understood with regards to mechanical loads originating from cleats: ➢ If cleats are very rigid, they perfectly fulfill their role by imposing the wave nodes of the deformed section of a warped stiffener inner flange. But then, the displacement of the flange at the cleat is nil. Consequently, the cleat is not the seat of any stress due to stiffener warping. ➢ If they are too flexible, they do not impose the wave nodes described above. In this case, stiffener warping would load the cleats. ➢ In one case like in the other, warping is prohibited. Therefore, cleats are parts which are not loaded, which does not mean that they are useless. It was thus checked in flight that the keel beam stabilization rod of the A300/310 was not the seat of any stress. However, the absence of this rod would make the keel beam incapable of bearing ultimate loads.

General information

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CLEATS

.2 DATA PREPARATION

In the remainder of this document the study is limited to a typical fuselage panel and frame warping. Transposing the method to other cases should not cause any problem.

V1-9

.2.1 MATERIAL DATA

The calculations carried out in this chapter only call for the following material values: - the elasticity modulus, - the elasticity modulus in compression. During the cleat checking process it may be necessary to recalculate frame warping. The material data simply needed for this calculation are not given here. Refer to Chapter V3-3 which specifically deals with this phenomenon.

V1-9

.2.2 GEOMETRICAL DATA

The main dimensions of the frame/stringer bays are known, being: - the inter-frame pitch: L - the inter-stringer pitch: W

Stringer

Frame Cleats

L

W

Figure V1-9 .2.2-1: Main dimensions of a bay (2 frames + 2 stringers)

The dimensions of the cleats are given in Figure V1-9 .2.2-2.

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CLEATS

Cleats l4 l3min

l5 e

l1

Stringer

Frame i + 1

l2

l3max

L Frame i

Figure V1-9 .2.2-2: Dimensions required for the cleat calculation

Remarks: - In Figure V1-9 .2.2-2, the dotted areas correspond to areas of cleats used to join the frame and the fuselage skin, to transfer pressurization loads, for example. The entire study in this chapter only concerns the stabilization role of the parts in bold lines. In this document, these parts are called "cleats" and will be considered as separate parts. - Dimension l1 is measured between the last fastener on the frame web and the fasteners of the cleat on the stringer. - l5 is defined by drawing two lines at ± 45° from the last fastener of the frame web and measuring the distance between the two intersection points of these lines and the edge of the cleat web (refer to Figure V1-9 .2.2-3). l4 45° 45° l5

Figure V1-9 .2.2-3: Definition of distance l5

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- The cleat calculated here comprises: 1) a flanged edge bearing on the frame web the width of which is constant (l2). 2) a web with height "l3" linearly evolutive between two values "l 3min" and "l 3max" 3) a constant thickness "e".

Also, it is necessary to know the cross section area and the moment of inertia of the stringer around z in relation to the centre of gravity of its cross section.

The specific dimensions of the frame inner flange are given in paragraph V1-3 .4.3.

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.3 CLEAT STIFFNESS

V1-9

.3.1 ASSUMPTIONS

V1-9 •

CLEATS

Frame cleats are fastened to the stringers. Therefore, the stringer is considered as participating in cleat stiffness. The portion of the cleat used for the frame - skin joint is not considered. The connection of the cleat to the frame is by means of several fasteners. Only the end fastener is taken into account by assuming that it is on the compression load-carrying part of the frame web (refer to Figure V1-9 .3.2-1). Also, bending of the stringer around y due to the offset of this load will be neglected. Stringers and frames are considered as straight beams the lengths of which are long in comparison with the size of their inner flanges and their webs.

V1-9

.3.2 DISPLACEMENT AND STIFFNESS

Cleat stiffness can be estimated considering the displacement of the point of application of load F on the last fastener. This displacement is the sum of the displacements due to: - cleat bending (∆lbending/Mz) - cleat rotation (∆lrotation/z) - cleat translation (∆ltranslation/x) - cleat flange in contact with the frame bending (∆lbending) Frame Cleat Frame

F ll Stringer

F

L

L

y x

Stringer

F Cleat y

ll

Cleat L

z x

Stringer W L

L y x

Figure V1-9 .3.2.-1: Mechanical model of the cleat/frame/stringer assembly

Cleat stiffness

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Displacement of the point of application of load F: ∆l ∆l = ∆lflexion/Mz + ∆lrotation/z + ∆ltranslation/x + ∆lbending

Eq. 1.9.3.2.1

The stiffness of the cleat is given, by definition, by the relationship: Stiffness: K

V1-9

K=

F ∆l

Eq. 1.9.3.2.2

.3.2.1 Cleat bending: ∆lbending/Mz F

Cleat

F A

∆lbending/Mz

A l1

l1 Stringer y y

L

x

L

x Section AA l2 l3 Figure V1-9 .3.2.1-1: Cleat bending Load F induces bending of the cleat around z. The cleat is assimilated to a beam clamped at the stringer. In compliance with the sign conventions given in Figure V1-9 .3.2.1-1, the bending moment at position y along the cleat is: Mfz(y) = F . (ll - y)

Eq. 1.9.3.2.1.1

The normal stress due to Mfz is: σy =

M fz ( y) . v( y) I Sz (y )

Eq. 1.9.3.2.1.2

with: v(y): x distance from the cross section CG to the fibre at position y ISz(y): moment of inertia of the cleat around z in relation to the centre of gravity of the cross section, at position y.

Cleat stiffness




V1-9 •

CLEATS

On Figure V1-9 .3.2.1-1 above, the dimension "l2" is the width of the flanged edge, whereas "l3" is the web height. Generally, 3l linearly change from one end of the cleat to the other: l3 = f (y). To calculate the deflection at the end of this evolutive cross section beam, an approximation can be made with a beam of equal strength. Remember that a beam of equal strength is a beam M fz for which the value σ0 = , i.e. the normal stress due to bending, is constant over its æ I Sz ö ç ÷ è v ø entire length (in particular refer to document ref. (6) Chapter X, paragraph 4.7). æI ö Here the bending moment linearly decreases with y, whereas the increasing ratio ç Sz ÷ è vø follows a polynomial law. In our case, the result is that the stress is not constant along the cleat but decreases towards its free end (refer to Figure V1-9 .3.2.1-2).

σ

Beam of equal strength

σ0 Cleat

ll

y

Figure V1-9 .3.2.1-2: Evolution of normal stress due to bending in the cleat. The applied stress decreases towards the end where F is applied. The deflection calculated using an assumption of a beam of equal strength will therefore be greater than the real deflection. The procedure consists in discretizing the beam into n segments of equal length ∆y along which the bending moment and the dimension v(y) are considered as being constant (refer to Figure V1-9 .3.2.1-3). Reference document (6) gives the equation of the deflection, in particular as a function of the stress calculated at the clamped end (y = 0). σ0 n M (y ) . å i = l fz i . ∆y Es F . v( y i ) (refer to Figure V1-9 .3.2.1-3)

∆lbending/Mz =

Cleat stiffness


Eq. 1.9.3.2.1.3



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CLEATS

with:

Mf0 æ I0 ö ç ÷ è v0 ø (stress at the fixed end)

σ0 = σ(y = 0) =

Eq. 1.9.3.2.1.4

Mf0: bending moment at y = 0, being Mf0 = F . ll I0: moment of inertia around z in relation to the cross section CG at y = 0 v0: x distance from the centre of gravity of the cross section to the end fibre at y = 0 Es: elasticity modulus of the cleat material Mfz (yi): bending moment at y = yi, being Mf (yi) = F . (ll - yi) v(yi): x distance from the centre of gravity of the cross section to the end fibre at y = yi l ∆y: length of a cleat segment, being ∆y = l n Clamped end (on the stringer) 1

F

∆y

2

3

…

i

…

n-1

n

ll

0

y

Cleat

yi

Figure V1-9 .3.2.1-3: Cleat Therefore, the deflection as a function of the applied load F is formulated as follows: ∆lbending/Mz =

F . l 2l . v 0 1 n (l − y i ) . . åi = l l Es .I0 n v(y i )

Eq. 1.9.3.2.1.5

A study on the deviations obtained for different values of "n" shows that in a special case where there is no flanged edge, i.e. when the cross section reduces to a variable height rectangle, the maximum deviation in relation to an infinite number of segments is obtained for n = 1. It is then only around 10%.

V1-9

.3.2.2 Cleat rotation: ∆lrotation/z

= L

Mf0 / 2

Mf0 / 2

Mf0

L

+ L

L

Figure V1-9 .3.2.2-1: Super-stringer bending

Cleat stiffness




V1-9 •

CLEATS

The clamping moment of the cleat on the stringer is: Mf0 = f.ll; Application of this moment bends the super-stiffener and therefore entails a rotation of the cleat assembly. Here the term "super-stringer" includes the stringer plus part of the skin. This is a super-stiffener as defined in Chapter V3-1. M .L with: The induced rotation is: θ = f 0 6. E l . I lz - L: inter-frame distance (refer to Figure V1-9 .1-1) - El: elasticity modulus of the stringer - Ilz: moment of inertia around z in relation to the super-stringer cross section CG. The displacement of the point of application of load F is therefore:

F

∆lrotation/z = ll.sinθ

∆lrotation/z

y x

Being: θ

æ F. l l . L ö ∆lrotation/z = ll.sin ç ÷ è 6. E l . I lz ø If angle θ remains small, sin θ ≈ θ

ll

L

L

Figure V1-9 .3.2.2-2: Cleat rotation Therefore: ∆lrotation/z ≈

V1-9

F. l l 2 . L 6. E l . I lz

Eq. 1.9.3.2.2.1

.3.2.3 Cleat translation: ∆ltranslation/x

The transverse load at the clamped end of the cleat on the stringer is F. Application of this load causes tension/compression of the super-stringer and therefore leads to translation of the whole cleat. ∆ltranslation/x Translation of the stringer and therefore of the cleat is: F. L 2. S l . E l Eq. 1.9.3.2.3.1 where Sl is the cross section of the stringer + the associated fuselage skin

y -F/2

-F/2

x

∆ltranslation/x =

Cleat stiffness


L Figure V1-9 .3.2.3-1: Cleat translation



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CLEATS

.3.2.4 Bending of the cleat flange in contact with the frame: ∆lbending

Load F induces bending around y of the cleat flange in contact with the frame. Section BB

F l5 l4

y

z x

Figure V1-9 .3.2.4-1: Bending of the flanged edge This flange is assimilated to a beam clamped at one of its ends (fastener side: refer to Figure V1-9 .3.2.4-2). Real structure Beam model

Figure V1-9 .3.2.4-2: Model of the equivalent beam for flange bending The maximum deflection of the beam is: ∆lbending =

F. l 34 3. E s . I bending

Eq. 1.9.3.2.4.1

where Ibending is the moment of inertia around y of section BB of the cleat. Remark: Remember that the cross section for bending calculation BB can be found by plotting two straight lines at 45° from the point of application of the load. Giving l5 = 2.l4. The compliance of the flange of a cleat, made of folded sheet (the thickness of which is limited) is relatively high and appreciably reduces the stiffness of the cleat.

Cleat stiffness




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CLEATS

.3.2.5 General equation of cleat stiffness: K

By carrying forward the equations 1.9.3.2.1.5, 1.9.3.2.2.1, 1.9.3.2.3.1 and 1.9.3.2.4.1 into the equation 1.9.3.2.1, the following is found: F . l 2l . v 0 F. l 2l . L F. l 34 n (l − y i ) F. L + + + ∆l = . å i =1 l 6. E l . I lz 2. S l . E l 3. E s . I bending n . Es .I0 v( y i )

Eq. 1.9.3.2.5.1

By carrying forward this equation to Eq. 1.9.3.2.2, the compliance is: l2 . v l2 .L l 34 n (l − y i ) 1 L = l 0 . åi =1 l + l + + 6. E l . I lz 2. S l . E l 3. E s . I bending K n . Es .I0 v( y i )

with: El: Ilz: Es: L: Ibending: ll: Mf0: I0: v0: Es: v(yi): ∆y: e:

S Eq. 1.9.3.2.5.2

elasticity modulus of the stringer moment of inertia around z in relation to the super-stringer cross section CG modulus of elasticity of the cleat inter-frame pitch moment of inertia around y of the cleat cross section BB (refer to paragraph V1-9 .3.2.4) distance between the end fastener on the frame and the row of fasteners on the stringer bending moment at y = 0, being Mf0 = F . ll moment of inertia around z in relation to the cross section CG at y = 0 y distance from the centre of gravity of the cross section to the end fibre at y = 0 elasticity modulus of the cleat material y distance of the centre of gravity of the cross section to the end fibre at y = yi length of a cleat segment cleat thickness Cleats y

E

S E N T

l4

l3min

I

l5 l1

Stringer

z l3 x

Frame i + 1

l2 G0 v0

L Frame i

Figure V1-9 .3.2.5-1: Dimensions required to calculate the cleat Note: G0 is the cross section CG at clamped end (y = 0)

Cleat stiffness


A


L


V1-9 •

CLEATS

.3.2.6 Calculation example y

l4

l5 l1

z l3 x

Frame i + 1

l2 G0 v0

L Frame i

Figure V1-9 .3.2.6-1: Reminder of the dimensions required for the cleat stiffness calculation

Cleat stiffness



Static stressing manual Cleat characteristics: F L ll mm N mm -1000 530 80 Es MPa 69000

i 1 2 3 4

v0 I0 mm4 mm 8920 20.25 yi mm 10 30 50 70

l3i mm 28.75 26.25 23.75 21.25

l2 mm 20 ISz mm4 16

l3 min mm 20

V1-9 • l3 max mm 30

l4 mm 12

l5 mm 24

CLEATS e mm 2

∆y mm 20

n 4

vi mm 19.525 18.057 16.564 15.039

Stringer characteristics: El Sl Ilz MPa mm4 mm² 73800 26540 177 Displacements: ∆lbending/z = ∆lrotation/z = ∆ltranslation/x = ∆lbending = ∆l =

-0,46 -0,29 -0,02 -0,52

mm mm mm mm

-1,30 mm

Refer to Eq. 3.9.3.2.1 or Eq. 3.9.3.2.5.1

Cleat stiffness

Refer to Eq. 3.9.3.2.1.5 Refer to Eq. 3.9.3.2.2.1 Refer to Eq. 3.9.3.2.3.1 Refer to Eq. 3.9.3.2.4.1 Stiffness: K = 772 N/mm Refer to Eq. 3.9.3.2.2 or Eq. 3.9.3.2.5.2




V1-9 •

CLEATS


Cleat stiffness




V1-9 •

V1-9

.4 CLEAT EFFECTIVENESS

V1-9

.4.1 STIFFNESS CRITERION

CLEATS

To determine the minimum stiffness that cleats must have, it is necessary to take the problem of frame stability into account. This problem is studied in detail in Chapter V1-3. The inner flange of the frame is considered as a continuous beam with a constant cross section supported by an elastic foundation, rigid supports at its ends, and "n" intermediate elastic supports of the same stiffness K, also equally spaced (all these supports represent the cleats). The cross section of the beam defined in this manner comprises the frame inner flange (refer to paragraph V1-3 .4). Ity is the moment of inertia of the cross section. The elastic foundation represents the stiffness of the frame web. It is quantified by parameter β, given in paragraph V1-3 .4. This stiffness is assimilated to additional inertia provided to the basic beam. It is thus possible to define an equivalent beam without elastic foundation the inertia of which is I' ty. Rn

Rn-1

R2

R1

P

P W

x

W

z

W

W

l

Figure V1-9 .4.1-1: Model of a beam under compression on elastic supports The first goal can be summarized as being that the intermediate supports (cleats) must not move if the beam buckles (frame warping). They will then be considered as being absolutely stiff supports. Let be K, the stiffness of the intermediate supports. It is proven that the supports can be considered as being absolutely stiff if K satisfies the equation: K > Kmini =

Pcr γ. W

(refer to V1-9 .7)

where: - Pcr is the critical Euler load for a beam section of length W - γ is a parameter depending on the number of cleats n (refer to document ref. (3), paragraph 2.6)

Cleat effectiveness




V1-9 •

CLEATS

n

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

20

γ

0.5000

0.3333

0.2929

0.2764

0.2679

0.2630

0.2599

0.2578

0.2563

0.2552

0.2543

0.2537

0.2532

0.2528

0.2524

0.2529

0.2519

0.2517

0.2514

0.5000

γ = γ(n)

0.4500

γ

0.4000 0.3500 0.3000 0.2500 0.2000

0

2

4

6

8

10 n

12

14

16

18

20

Figure V1-9 .4.1-2: Evolution of parameter γ as a function of the number of cleats If there are not many cleats (n<5), it may be useful to take the values of γ indicated in the table. Otherwise, the asymptotic value of the curve γ = 0.25 has to be used (this is the case for fuselage frames). The minimum allowable stiffness of the cleat is thus maximized. It can be noted that the greater the number of cleats, the higher their stiffness has to be to oppose the same buckling force. This can be explained by the fact that on the beam of length l, as n increases then W decreases and therefore this is a short beam case. Under these conditions, as n increases, Pcr will increase and thus the resisting load to be provided during buckling in the event of support displacement, which directly depends on stiffener K, has to be high. The new inertia I' ty is defined by formulating the equation of the critical Euler load for the beam of length W in two forms: Critical Euler load for a continuous beam on an elastic foundation: ö π 2 . E cc . I ty æ 2 β. W 4 ç ÷ Pcr = p . + 2 4 2 ç p . π . E cc . I ty ÷ø W è

Eq. 1.9.4.1.1

with: W: Ecc: Ity: β: p:

length of a span between two supports elasticity modulus in compression of the frame material minimum moment of inertia of the frame inner flange (refer to paragraph V1-3 .4) elastic foundation modulus (refer to paragraph V1-3 .4) number of half waves over length W

Critical Euler load for a continuous beam on two supports: π 2 . E cc . I'ty Pcr = W2

Cleat effectiveness


Eq. 1.9.4.1.2



V1-9 •

CLEATS

Concerning the beam of length W, the two equations of Pcr given below may be equalized. The inertia of the equivalent beam which comprises the frame inner flange plus the effect of the elastic foundation is therefore: æ 2 ö β. W 4 ÷ I' p + ty = Ity . ç ç p 2 . π 4 . E cc . I ty ÷ø è

Eq. 1.9.4.1.3

Consequently, the stiffness criterion ifγ = 0.25 becomes: K > Kmini =

4Pcr W

With: Pcr =

π 2 . E cc . I'ty

W3 Therefore the stiffness criterion may also be formulated as follows:

K>

4 π 2 . E cc . I'ty

Eq. 1.9.4.1.4

W3

Remark: Cleat and frame inner flange sizing satisfying the stiffness criterion may be linked by the equation: K. W 3 Ecc.I' ty = 4π 2

Cleat effectiveness




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CLEATS

.4.2 NON-RIGID CLEATS

If the stiffness criterion is not satisfied and the cleat is not resized, it is possible to accept that a cleat is "flexible" in relation to the frame inner flange. The problem then becomes one of the buckling of a beam on elastic supports, studied in particular in reference document (3), paragraph 2.6. The general formulation of the critical Euler load for the stiffener involved on elastic supports is: Pcr =

c . π 2 . E cc . I'ty W2

with: Ecc: I' ty: W: c . π²:

elasticity modulus in compression of the frame material minimum inertia moment of the frame inner flange around y in relation to its centre of gravity, taking the elastic foundation into account beam length (inter-stringer pitch) dimensionless factor determined using the chart and partly depending on the stiffness of the intermediate elastic supports (refer to below).

To calculate c . π², a chart has be used with the entry on the abscissa line being a noted dimensionless quantity (refer to Figure V1-9 .4.2-2). K. W 3 for a beam of length "W"where K is the stiffness of an intermediate elastic E cc . I'ty support. The coefficient "cπ²" for a beam of length "W" is obtained from the charts given in Figure V1-9 .4.2-2. This chart is valid for a number of supports tending towards the infinite but it can be accepted that it remains valid in all cases.

10 cπ² 5 0 0

20

40 K. W 3 E cc . I'ty

60

Figure V1-9 .4.2-1: Determination of coefficient cπ²

Cleat effectiveness




V1-9 •

CLEATS

K. W 3 > 40, the following may be taken into account c ≈ 1 and c.π² ≈ π² E cc . I'ty This curve may be approximated by a polynomial interpolation. Putting down:

For

X=

K. W 3 E cc . I'ty

For X < 40, there is then;: -2 . X2 + 4.2197.10-3 . X3 - 9.5732.10-5 . X4 + 8.2719.10-7 . X5 cπ² = 0.53591 + 1.1737 . X - 9.1984.10 The maximum deviation between the values read on the chart and those obtained by this equation is 18.8% for X ≈ 1.25. It must be remembered that this error overlays the interpretation error made when reading the curve. Putting down λ =

Pcr =

1 , the end fixity factor, the new critical Euler load is then: c

π 2 . E cc . I'ty

(λ . W ) 2

Thus, if the cleats are not rigid with regard to a beam of length W, they become rigid for a beam of length λW. Once the new value of the critical load Pcr has been determined, the allowable warping stress ~ , refer to Chapter V1-3) must be recalculated and checked to make sure that it is still (σ dr greater than the applied stress. Using the notations developed above, the allowable warping stress is: ~ = Pcr < σ ~ σ dr dr initial St

Eq. 1.9.4.2.5

with: St: cross section area of the frame inner flange Calculation details, especially a possible plasticity correction, are given in Chapter V1-3.

Cleat effectiveness




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Cleat effectiveness




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CLEATS

.5 CHECKING PROCEDURE

E

BEGINNING Calculation of cleat flexibility: l2 . v l2 .L l 34 n (l − y i ) 1 L = l 0 . åi =1 l + l + + 6. E l . I lz 2. S l . E l 3. E s . I bending K n . Es .I0 v( y i ) æ 2 ö β. W Calculation of corrected inertia: I' ty = Ity . ç ç p + p 2 . π 4 . E . I ÷÷ è cc ty ø

Eq. 1.9.3.2.5.2

S

4

Calculation of the required minimum stiffness: Kmin = no yes

Eq. 1.9.4.1.2

4 π 2 . E cc . I'ty W

3

Eq. 1.9.4.2.3

yes

K > Kmini

E

no

Resize the cleat?

Determine c.π² from

S

K. W 3 E cc . I'ty

N

Eq. 1.9.4.2.1 + Fig. V1-9 .4.2-1: 10

T

cπ² 5 0 0

20

40

60

K. W 3 E cc . I'ty

Calculation of the new critical load: Pcr =

c. π 2 . E cc . I'ty

I Eq. 1.9.4.2.2

W2

~ : refer to paragraph V3-3 for the initial calculation, or σ ~ = Pcr Calculation of the new σ dr dr St Eq. 1.9.4.2.3 yes no

Resize the frame?

War ping?

no

L

yes END

Checking procedure


A



V1-9 •

CLEATS

The calculation of the stiffness specific to cleats is discussed in paragraph V1-9 .3. The calculation of the stiffness criterion is given in paragraph V1-9 .4. If the allowable stress becomes insufficient, it may simply be preferable to resize the frame inner flange. This avoids defining excessive cleat geometries as indicated in paragraph V1-9 .1.

Checking procedure




V1-9 •

CLEATS

6-CALCULATION EXAMPLE

The example of the frame discussed below first of all comprises the calculation of the initial warping stress as indicated in paragraph V1-3 .4.8. Here, part of the same notations are partly reused, here being: et

bt Z CdGflange

ea ha

W

Y X Figure V1-9 .6-1: Dimensions required to calculate frame warping - Frame material: - Ultimate tension allowable stress: - Elasticity modulus in compression: - Compression yield stress: - Inner flange thickness: - Inner flange width: - Web thickness: - Web height:

7075 T7351 σr = 495 Mpa Ecc = 73800 Mpa σc0.2 = 420 Mpa et = 4 mm bt = 16 mm e a = 2 mm h a = 110 mm

The inter-frame pitch L in example V1-3 .4.8 is replaced by the inter-stringer pitch. - Inter-stringer pitch: W = 176 mm

Calculation example




V1-9 •

CLEATS

Inner flange section: St = et . bt = 4 x 16 = 64 mm² Moment of inertia of the inner flange at its cross section CG and its principal axes: (refer to V3-3 .4.3) e . b 3 4 x 16 3 = 1365,3 mm4 Ity = t t = 12 12 Stiffness of the elastic foundation: E β = cc 4

3

3 æe ö 73800 æ 2 ö .ç a ÷ = .ç ÷ = 0.111 MPa è 110 ø 4 è ha ø

Calculation of p0 (critical number of half waves): β W 176 0.111 = .4 .4 E cc . I ty 73800 x 1365,3 π π

p0 =

= 0.323

Giving: p1 = 0 p2 = 1 Calculation of the initial critical stress: ~ (p ) = 2 . E . I . β = 2 x 73800 x 1365,3 x 0.111 = 105 MPa σ 0 dr cc ty St 64

As the number of half waves is necessarily equal to or greater than 1, the following cannot exist p = 0. Consequently, p = 1 and: ~ ( p ) éæ 1 ö 2 ù 105 éæ 1 ö 2 ù σ 2 0 dr ~ . êç ÷ + ( p 0 ) ú = x êç ÷ + (0.323) 2 ú = 509 MPa (σ dr ) initial = 2 2 ëè 0.323 ø úû êëè p 0 ø û The inertia corrected as a function of the elastic foundation is: æ 2 ö æ 2 ö 0.111 x 176 4 β. W 4 4 ç ÷ I' = I . p + = 1365 3 x 1 + , ç ÷ = 1380 mm ty ty ç 2 4 2 4 ÷ p . π . E cc . I ty ø 1 x π x 73800 x 1365,3 ø è è

Therefore, the critical compression load is: Pcr =

π 2 . E cc . I'ty W2

Calculation example

π 2 x 73800 x 1380 = = 32450 N 176 2




V1-9 •

CLEATS

The minimum inertia required for the cleat to be considered as being rigid is finally: Kmin =

4 Pcr 4 x 32450 = 737 N/mm W 176

According to the cleat stiffness calculation discussed in paragraph V1-9 .3.2.6, there is K = 772 N/mm. Therefore, K > Kmin, the cleat may be considered as a rigid support.

Now assume that K = 600 N/mm. Therefore there is K < K min and the cleat is not rigid. The cleat is not resized. It is necessary to determine the coefficient " c.π²": K. W3 600 x 176 3 = = 32 Þ c . π² ≈ 9.54 E cc . I'ty 73800 x 1380 The critical compression load therefore becomes: Pcr =

c . π 2 . E cc . I'ty W

2

=

9.5 x 73800 x 1380 = 31234 N 176 2

Now the critical warping stress has to be recalculated: ~ = Pcr = 31234 = 488 MPa σ dr 64 St

(to be compared to the 509 Mpa initially obtained)

Note that a plasticity correction is required to obtain the real warping stress. This aspect of the calculation is outside the framework of this chapter and therefore not discussed here (refer to Chapter V1-3).

Calculation example




V1-9 •

CLEATS


Calculation example



Technical Manual MTS 004 Iss. B External distribution authorised:

YES

X

NO


1 4

5 4

2 1 2

5

Volume 2


Purpose Scope

IT tool supporting this Manual

Contents

Document Manager

Strength of metallic materials All programmes.

Not applicable.

V2 - 1 V2 - 2 V2 - 3 V2 - 4 V2 - 5 V2 - 6


Characteristics of sections Behaviour of materials Strain measurement Bending of beams Columns Beams-columns

Validation

Name: J. HUET

Name: JF. IMBERT Function: Deputy Department Group Leader Dept code: BTE/CC/A Date: 11/99 Signature

This document is the property of AEROSPATIALE MATRA AIRBUS, no part of it shall be reproduced or transmitted without authorisation of AEROSPATIALE MATRA AIRBUS and its contents shall not be disclosed. © AEROSPATIALE MATRA AIRBUS - 1999

4page 1


1.


MTS 004 Ind. B

4page 2

Static stress manual, metallic materials - Annex

Reference documents


Abbreviations




Definitions


Highlights

Issue

Date

Pages modified

A

02/98

V2 - 4

New document.

B

11/99

V2 - 4

Changes as per table page V2-4.i

V2 - 5

New chapter.

V2 - 6

New chapter.


Justification of the changes made

MTS 004 Iss. B

4Ann. page



List of approval

Dept. code

Function

BTE/CC/CM

Name / First name


Key words

Calcul

Bibliography

Néant

Signature

CAZET G.

Distribution list

Dept. code

Function


BQP/TE

Archives Diderot

SIBADE Alain

BQP/TE


SIBADE Alain


BOUTET Fernand

BTE/SM/MG



MTS 004 Iss. B

page IG1

SSM V2-4 • BENDING OF BEAMS CONTENTS issue

date

revision

V2-4 BENDING OF BEAMS

1

5/1999

New § 5 and 6

V2-4 1 INTRODUCTION

0

6/1997

Creation

V2-4 2 STRAIGHT BEAMS

1

9/1999

Correction p.2-8

V2-4 3 BEAMS WITH HIGH CURVATURES

0

6/1997

Creation

V2-4 4 BEAMS WITH VARIABLE HEIGHTS

0

6/1997

Creation

V2-4 5 CALCULATING STRESSES IN STRAIGHT BEAMS

0

5/1999

Creation

V2-4 6 CALCULATING STRESSES IN CURVED BEAMS

0

5/1999

Creation

V2-4 1 INTRODUCTION

1-1

V2-4 2 STRAIGHT BEAMS

2-1

V2-4 V2-4 V2-4 V2-4

2.1 PLANE BENDING FORMULA 2.1.1 Conditions for use 2.1.2 Notations and Conventions 2.1.3 Form

V2-4 2.2 EXAMPLES

2-1 2-1 2-2 2-3 2-17

V2-4 3 BEAMS WITH HIGH CURVATURES V2-4 3.1 GENERAL V2-4 3.2 THEORETICAL BASIS V2-4 3.3 PRACTICAL FORMULAS

3-1 3-1 3-1 3-2

V2-4 4 BEAMS WITH VARIABLE HEIGHTS V2-4 4.1 CORRECTION DUE TO "M" V2-4 4.2 CORRECTION DUE TO "N" V2-4 4.3 PRACTICAL FORMULAS AND COMMENTS

V2-4 5 CALCULATING STRESSES IN STRAIGHT BEAMS

4-1 4-1 4-1 4-2

5-1

V2-4 5.1 LINEAR ELASTICITY V2-4 5.1.1 Plane bending V2-4 5.1.2 Deviated bending V2-4 5.1.3 Composite beams

5-1 5-1 5-1 5-4

V2-4 5.2 PLASTICITY

5-6


Contents

page •i

SSM V2-4 • BENDING OF BEAMS

V2-4 5.2.1 Plane bending V2-4 5.2.2 Deviated bending

5-6 5-16

V2-4 6 CALCULATING STRESSES IN CURVED BEAMS

6-1

V2-4 6.1 SYMMETRICAL SECTIONS V2-4 6.1.1 Introduction V2-4 6.1.2 Plane elastic bending of solid sections V2-4 6.1.3 Plane elastic bending of thin profile sections

6-1 6-1 6-1 6-2

V2-4 6.2 ASYMMETRICAL THIN PROFILE SECTIONS V2-4 6.2.2 Introduction V2-4 6.2.2 Calculating load-carrying section V2-4 6.2.3 Example

6-8 6-8 6-8 6-9

page •ii © AEROSPATIALE - 1999

Contents

Issue 1

SSM V2-4 • BENDING OF BEAMS V2-4 1

INTRODUCTION

The purpose of Chapter V2-4 is two-fold: - determine internal loads and related distortion values, - find stress values.

Internal loads

Tables provide algebraic solutions for internal loads and distortion in different cases of straight beams carrying loads in the elastic range. This concerns simple cases based on the plane bending theory which, when used with the principle of superposition makes it possible to solve most problems or to obtain a satisfactory estimate in the most complex cases (refer to the assumptions used in "straight beams"). Stresses are then determined using conventional formulas associated with the plane bending theory. These simple formulas are given here.

Stresses

The stress calculation is detailed for beams with a geometry and/or a loading mode, especially the direction or the point of load application, prohibiting the use of partial or complete results obtained from the plane bending theory. For example, this is the case of beams with a non-negligible radius of curvature ("curved beams") or beams with variable heights. In these particular cases, stress determination may involve an internal load distribution that is very different from the one obtained with the plane bending theory.


Introduction

page V2-4•1-1



page • V2-4•1-2 © AEROSPATIALE - 1997

Introduction

Issue 0


STRAIGHT BEAMS

V2-4 2.1 PLANE BENDING FORMULA For various loading cases, the following tables give the expressions to be used to obtain, at all points along the span of a beam, the value of the resulting loads M (bending moment) and T (shear force) and the characteristics of the deflected beam θ (slope) and y (deflection). The value of the reactions and the value of the distortion at each end are called "end values". Specific values, determined using specific geometrical or loading data are called "remarkable values" (for example: minimum and/or maximum values of M…). Each table covers a loading type case in two steps: the header of the table gives the general expressions, then the end values and the remarkable values are given for each of type of boundary conditions. The last two cases concern the imposed point displacement and rotation cases.

V2-4 2.1.1

Conditions for use

The restrictions given below are directly related to the conditions required to qualify the load-carrying mode of an plane bending beam, these conditions being: - strain of the beam in a plane common to the loading plane. This plane defines the bending plane, - no warping and maintaining of the cross sections after strain in a plane normal to the neutral fibre, - no rotation of sections around a longitudinal axis guaranteeing a pure bending state, i.e. without torsion; Restrictive conditions: 1) the beam is long (length relatively long in relation to transverse dimensions); 2) the beam may have a solid or thin section but must be stable; 3) the forces applied act in the same longitudinal plane; 4) this plane includes the shear centre of the cross sections (the shear centre is in the longitudinal plane of symmetry, if there is one. It is comparable with the centre of gravity for solid sections and for thin closed sections; 5) the cross sections are constant or vary little along the beam; 6) the beam is straight or slightly curved (high radius of curvature in relation to transverse dimensions); 7) the material is homogeneous, isotropic with a linear elastic range and is stressed in this range; 8) the small displacement assumption is used. Observance of these conditions makes it possible to take into account the deflection at any point due to the bending moment M (complementary deflection due to T is not taken into account), disregard neutral fibre elongation and therefore variations in the beam length and makes the stress normal to a point proportional to the distance separating it from the neutral fibre. On the contrary, examine the beam taking the following phenomena into account given respectively for the conditions above when not fulfilled: 1) the interdependency of the phenomena in the three directions as described by the elasticity theory and which is disregarded in plane bending, may be high when the beam slenderness is low (Poisson effect); 2) general instability due to the longitudinal components of the applied forces (refer to Chapters 5 and 6); 3) the neutral fibres of the various sections are not included in a single plane. This phenomenon generates torsion (for the normal stress calculation it is acceptable however to break loading down into two bending planes); 4) essentially for thin open sections: lateral buckling, in particular causing an increase in deflections and contributing to the torsion thus induced in the normal stress calculation.


Straight beams

page V2-4•2-1

SSM V2-4 • BENDING OF BEAMS 5) Correction of the shear load in the case of variable height beams; 6) Nonlinear distribution of stresses for a given section and complementary deflection due to non-negligible T in the case of high initial curvature; 7) Behaviour specific to composite materials or to the plastic range; 8) In general, the problem requires processing by the finite element method.

V2-4 2.1.2

Notations and Conventions

F: applied point load, expressed in N. p: applied load per unit length, expressed in N/mm. Mo: applied point moment, expressed in mmN.

The terms F, p, Mo, θo, ∆o designate algebraic values.

θo: imposed rotation, expressed in rd. ∆o: imposed deflection, expressed in mm.

All forces, moments and displacements applied are positive and oriented as shown in the tables.

!

Remark: the minimum and maximum algebraic values given in the tables are positive values, i.e., as drawn. For negative values, the expressions marked MAX and MIN respectively indicate the minimum algebraic and maximum algebraic value. RA (RB): reaction at bearing point A (B), expressed in N, positive upwards. MA (MB): end moment A (B), expressed in mmN, positive when it compresses the top fibres. M: bending moment, expressed in mmN, positive when it compresses the top fibres. T: shear force, expressed in N, positive upwards when it concerns the left end of a portion of a beam.

θ: slope, expressed in rd, positive direction given by the axis-system. y: deflection, expressed in mm, positive direction given by the axis-system.

All values positive

y M z

T

M+∆M

θ

θ

y x

T+∆T This convention implies: ΣFLEFT = T ; ΣFRIGHT = - T ; ΣMT(FLEFT) = M ; ΣMT(FRIGHT) = M ; in particular: RA = T (x = 0) ; RB = - T (x = l) ; MA = M (x = 0) ; MB = M (x = l) The stresses obtained are given in MPa in the unit system above. The formulas given at the top of each table show the binary function áx - año. It is used alone or combined with ordinary algebraic functions and is defined as follows: Let x be the abscissa of a cross section, δ the location parameter of a load component: if x < δ, then áx - δño = 0 ;

y

δ

if x > δ, then áx - δñ o = 1 ;

F

at x=δ, the function áx - δño may be undefined as, for example, the shear force T at position where a point load is applied: x

page V2-4•2-2 © AEROSPATIALE - 1999

x

by extension, áx - δñn is interpreted as the product (x - δ)n áx - δño. Straight beams

Issue 1


V2-4 2.1.3

Form

The form is divided into five tables, each representing a type of load: 1_concentrated load, 2_distributed load, 3_point moment, 4_imposed deflection, 5_imposed rotation. The formulas used to calculate, at any point along the span, the shear force and the bending moment values as regards to the resulting forces, and the slope and deflection values as regards to the distorted part, are called "General Expressions". Each table is divided into six parts and each part represents a combination of beam boundary conditions: a_one free end, one clamped end, b_one end blocked in rotation, one clamped end, c_one end on a single bearing point, one clamped end, d_two clamped ends, e_two free ends, f_one end blocked in rotation, one end on a single bearing point. The reactions and the slope and deflection values at ends are called "End Values". Some significant results are given (not exhaustively) under the title "Remarkable Values". A complete case is processed by combining the results of one line of the table with the results in the header.


Straight beams

page V2-4•2-3

SSM V2-4 • BENDING OF BEAMS 1 - CONCENTRATED LOADS General expressions

y

o

F

MB

a yA

Shear force: T = RA - Fáx - añ Bending moment: M = MA + RAx - Fáx - añ

θB x

MA

MA x RA x 2 F + − áx - añ2 EI 2EI 2EI M A x 2 RA x 3 F Deflection: y = yA + θAx + + − áx - añ3 2EI 6EI 6EI Slope: θ = θA +

RB

θA l

RA o

o

o

o

If x < a thenáx - añ = 0 ; if x > a then áx - añ = 1 = ; by extension, áx - añ is interpreted as the product (x - a)n áx - añ .

end on a single bearing point

free end 1-a F a

A

end blocked in rotation

clamped end

RA = 0

MA = 0

F θA = (1 - a)2 2EI

RB = F

MB = - F(l - a)

θB = 0

End values -F 3 2 yA = (2l - 3l a + a3) 6EI yB = 0

B Remarkable values

MMIN = MB ; if a = 0, MMIN = - Fl

Fl 2 θMAX = θA ; if a = 0 θMAX = 2EI

yMIN = yA ; if a = 0, yMIN =

1-b

- Fl 3 3EI End values

F

a

B

A

RA = 0

MA =

F (l − a) 2 l

RB = F

MB =

- F (l 2 − a 2 ) 2 l

2

-F (l - a)2 (l + 2a) 12EI

θA = 0

yA =

θB = 0

yB = 0 Remarkable values

Fl ì ï M MAX = M A ; if a = 0, M MAX = 2 M í − Fl ï M MIN = M B ; if a = 0, M MIN = 2 î

yMIN = yA ; if a = 0, yMIN =

- Fl 3 12EI

1-c

End values

a

RA =

F (l − a) (2l + a) 2 l3

MA = 0

RB =

F a(3l 2 − a 2 ) 2 l3

MB =

2

F

B

A

- F a(l 2 − a 2 ) 2 l2

θA =

−F a (l - a)2 4EI l

θB = 0

yA = 0 yB = 0

Remarkable values

ì F a(l − a) (2l + a) ï for x = a ; if a = 0.366 l, M MAX = 0.174 Fl M í M MAX = 2 l3 ïî M MIN = M B ; if a = 0.5773 l, M MIN = − 0.1924 Fl 2

1/ 2 1/ 2 ì æ a ö æ a ö −F ïif a > 0.414 l: y MIN = a( l − a ) 2 ç for x = lç ÷ ÷ 6 EI è 2l + a ø è 2l + a ø ï ïï l (l 2 + a 2 ) − F a( l 2 − a 2 ) 3 y íif a < 0.414 l: y MIN = for x = 3 EI ( 3l 2 − a 2 ) 2 3l 2 − a 2 ï 3 ï −3 Fl for x = a ïif a = 0.414 l: y MIN = − 9.8 x 10 EI ïî


Straight beams

Issue 1

SSM V2-4 • BENDING OF BEAMS 1 - CONCENTRATED LOADS

1-d

End values

F

a

A

RA = F

(1 − a) 2 (1 + 2a) l3

MA = - F

a(1 − a) 2 l2

θA = 0

yA = 0

RB = F

a 2 (3l − 2a) l3

MB = - F

a 2 (1 − a) l2

θB = 0

yB = 0

B

Remarkable values ì ï M MAX M í ï if a < ïî

yMIN

a (l − a) l Fl for x = a ; if a = , M MAX = 3 l 2 8 l l : M MIN = M A ; if a = , M MIN = − 0.1481 Fl 2 3 = 2F

2

2

ì − 2Fa 3 (l − a) 2 l 2al for x = ïï If a < : yMIN = 2 3EI (l + 2a) 2 1 + 2a í 3 ï if a = l : yMIN = − Fl for x = a ïî 2 192 EI

1-e

End values

a

F

A

RA = F

(1 − a) l

MA = 0

θA =

− F a(2l − a) (l − a) 6EI l

yA = 0

RB = F

a l

MB = 0

θB =

F a(l 2 − a 2 ) 6EI l

yB = 0

B

Remarkable values

MMAX = F

yMIN

a(l − a) l Fl for x = a ; if a = , MMAX = l 2 4

3/2 1/2 ì æ l 2 − a2 ö l − F a æ l 2 − a2 ö ï If a < : yMIN = for x = l − ç ç ÷ ÷ ï 2 3EI l è 3 ø è 3 ø í l − Fl 3 ï ïî if a = 2 : yMIN = 48EI for x = a

1-f

End values

a

F

A

B

RA = 0

MA = F(l - a)

θA = 0

RB = F

MB = 0

θB =

−F (l - a) (2l2 + 2al - a2) 6EI


MMAX = MA for 0 ≤ x < a ; if a = 0, MMAX = Fl Fl 2 θMAX = θB ; if a = 0, θMAX = 2EI − Fl 3 yMIN = yA ; if a = 0, yMIN = 3EI


F 2 2 (l - a ) 2EI

yA =

Straight beams

page V2-4•2-5

SSM V2-4 • BENDING OF BEAMS 2 - DISTRIBUTED LOAD

pl

y

pa MB

a MA

yA

θB x

Slope: θ = θA +

RB

θA l

RA

General expressions pl − pa 2 Shear force: T = RA - pa áx - añ x−a 2(l − a) p p − pa 2 3 Bending moment: M = MA + RAx - a x − a − l x−a 2 6(l − a)

M A x RA x 2 p + − a x−a EI 2EI 6EI

Deflection: y = yA + θAx +

3

−

pl − pa x−a 24EI(l − a)

4

p pl − pa MA x 2 RA x 3 4 + − a x−a − x−a 2EI 6EI 24EI 120EI(l − a)

5

If x < a thenáx - añ0 = 0 ; if x > a then áx - añ0 = 1 ; by extension,áx - añn is interpreted as the product (x - a)n áx - añ0. end on a single end blocked in clamped end free end rotation bearing point

2-a pl a

pa

RA = 0 − pa 24EI p + RB = a 2

yA =

A

B

3p + pl MA = 0 θA = a (l - a)3 24EI p − pa (l - a)3 (3l + a) - l (l - a)3 (4l + a) 120EI pl 2pa + pl (l - a) MB = (l - a)2 6

2-b

a

pa

A

RA = 0 − pa 24EI p + RB = a 2

yA = B

3pa + pl

yB = 0

− pl 4 8EI − pl l 4 30EI − 11pal 4 120EI

End values

(l - a) θA = 0 24I p − pa (l - a)3 (l + a) - l θB = 0 yB = 0 (l - a)3 (3l + 2a) 240EI p − pa pl − pa (l - a)2 (2l + a) - l (l - a)2 (3l + a) (l - a) MB = 6l 24I MA =

Case of an evenly distributed load on l, i.e. a = 0 and pa = pl = p: ì pl 2 ïï MMAX = M A = − pl 4 6 M í y = y = MIN A 2 24EI ï MMIN = M B = − pl ïî 3 Case of an evenly increasing load on l, i.e. a = 0 and pa = 0: ì pl l 2 ïï MMAX = M A = − pl l 4 24 M í yMIN = yA = 2 80EI ï MMIN = M B = − pl l ïî 8 Case of an evenly decreasing load on l, i.e. a = 0 and pl = 0: ì pal 2 ïï MMAX = M A = − 7 pal 4 8 M í yMIN = yA = 2 240EI ï MMIN = M B = − 5 pal ïî 24


θB = 0

Remarkable values

Case of an evenly distributed load on l, i.e. a = 0 and pa = pl = p: − pl 2 pl 3 MMIN = MB = θMAX = θA = yMIN = yA = 2 6EI Case of an evenly increasing load on l, i.e. a = 0 and pa = 0: − pl l 2 p l3 MMIN = MB = θMAX = θA = l yMIN = yA = 6 24EI Case of an evenly decreasing load on l, i.e. a = 0 and pl = 0: − pal 2 p l3 MMIN = MB = θMAX = θA = a yMIN = yA = 3 8EI

pl

End values

Straight beams

3

Remarkable values

Issue 1

SSM V2-4 • BENDING OF BEAMS 2 - DISTRIBUTED LOAD

2-c

End values pa (l − a )3 ( 3l + a ) pl − pa (l − a ) 3 ( 4l + a ) RA = + MA = 0 pa l3 l3 8 40 a − pa ( l − a ) 3 (l + 3a ) pl − pa (l − a ) 3 ( 2l + 3a ) θA = yA = 0 − 48 EI l 240 EI l p + pl 2 pa + pl A B (l - a) MB = RAl (l - a)2 RB = - RA + a 2 6 θB = 0 yB = 0 Remarkable values Case of an evenly distributed load on l, i.e. a = 0 and pa = pl = p: 9 3 3 ì − pl 3 MMAX = pl 2 for x = l RA = pl =θA = θ MIN ï ï 128 8 8 48 EI M í 4 5 − pl 2 −3 pl ï MMIN = M B = RB = pl x 5 . 4 10 y = − for x = 0.4215 l MIN ïî 8 8 EI Case of an evenly increasing load on l, i.e. a = 0 and pa = 0:

pl l 10 2p l RA = l 5 RA =

pl

ì M MAX = 2.98 x 10 −2 pl l 2 for x = 0.4472 l ï M í − pl l 2 M = M = B ïî MIN 15

θ MIN = θ A =

− pl l 3 120 EI

y MIN = − 2.39 x 10 −3

pl l 4 for x = 0.4472 l EI

Case of an evenly decreasing load on l, i.e. a = 0 and pl = 0:

11 pal 40 9 RB = pal 40 RA =

ì M MAX = 4.22 x 10 −2 p a l 2 for x = 0.329 l ï M í −7 2 ïî M MIN = M B = 120 p a l

2-d

θ MIN = θ A =

− pa l 3 80 EI

y MIN = − 3.04 x 10 −3

pl l 4 for x = 0.4025 l EI End values

pa (l − a ) 3 (l + a ) pl − pa (l − a ) 3 ( 3l + 2a ) + 2 20 l3 l3 pa 3 a − pa (l − a ) (l + 3a ) pl − pa (l − a ) 3 ( 2l + 3a ) θA = 0 yA = 0 − MA = 12 60 l2 l2 p + pl 2 pa + pl RB = - RA + a (l - a) MB = RAl + MA (l - a)2 A B 2 6 θB = 0 yB = 0 Remarkable values Case of an evenly distributed load on l, i.e. a = 0 and pa = pl = p: ì pl 2 l for x = ïï M MAX = − pl 4 l 12 2 for x = M í yMIN = 2 − pl 384 EI 2 ï M MIN = M A = M B = ïî 12 Case of an evenly increasing load on l, i.e. a = 0 and pa = 0: 3 − pl l 2 7 RA = MA = RB = pll pll 30 20 20 ì M MAX = 2.15 x 10 −2 pl l 2 for x = 0.548 l p l4 ï M í yMIN = 1.309 x 10-3 l for x = 0.525 l − pl l 2 EI ïî M MIN = M B = 20 pl


RA =

Straight beams

page V2-4•2-7

SSM V2-4 • BENDING OF BEAMS 2 - DISTRIBUTED LOAD CONTINUED y

pl pa MB

a MA

yA

θB x

Slope: θ = θA +

RB

θA

General expressions pl − pa 2 áx - añ Shear force: T = RA - pa áx - añ 2(l − a ) p p − pa Bending moment: M = MA + RAx - a áx - añ2 - l áx - añ3 2 6(l − a )

l

RA

M A x RA x 2 p + − a x−a 2EI 6 EI EI

3

−

pl − pa x−a 24 EI (l − a )

4

MAx2 RAx3 p pl − pa 4 5 + − a x−a − x−a 2EI 6EI 24EI 120EI(l − a) If x < a thenáx - añ0 = 0 ; if x > a then áx - añ0 = 1 ; by extension,áx - añn is interpreted as the product (x - a)n áx - añ0. end on a single end blocked in clamped end free end rotation bearing point

Deflection: y = yA + θAx +

2-e

End values 2 pa + pl 2 (l - a) MA = 0 yA = 0 RA = pl 6 − pa (l − a ) 2 (l 2 + 2al − a 2 ) pl − pa (l − a ) 2 (7 l 2 + 6 al − 3a 2 ) θA = − pa a l l 24 EI 360 EI 2 pa + pl pa + pl 2 (l − a ) + (l − a ) MB = 0 yB = 0 RB = − 6 2 A B pa (l 2 − a 2 ) 2 pl − pa (l − a ) 2 ( 8l 2 + 9 al + 3a 2 ) θB = + l l 24 EI 360 EI Remarkable values Case of an evenly distributed load on l, i.e. a = 0 and pa = pl = p: pl pl 2 − 5 pl 4 l pl 3 l for x = for x = MMAX = R A = RB = θMAX = θB = yMIN = 384 EI 2 2 8 24 EI 2 Case of an evenly increasing load on l, i.e. a = 0 and pa = 0: pl − 7 pl l 3 MMAX = 6.41 x 10-2 pll2 for x = 0.5773 l θA = RA = l 360 EI 6 4 pl l pl l 3 p l yMIN = - 6.53 x 10-3 l for x = 0.5195l RB = θB = EI 3 45 EI

2-f pl a

pa

A

RA = 0 − pa 24 EI P + RB = a 2 yB = 0 yA =

B

End values 2 pa + pl 2 (l - a) MA = θA = 0 6 p − pa (l - a)2 (9l2 + 2al - a2) (l - a)2 (5l2 + 2al - a2) - l 120 EI pl p p −p (l - a) MB = 0 θB = a (l - a)2 (2l + a) + l a (l - a)2 (3l + a) 6EI 24EI Remarkable values

Case of an evenly distributed load on l, i.e. a = 0 and pa = pl = p: pl 2 pl 3 yMIN = yA = MMAX = MA = θMAX = θB = 2 3 EI Case of an evenly increasing load on l, i.e. a = 0 and pa = 0 p l2 p l3 yMIN = yA = MMAX = MA = l θMAX = θB = l 6 8 EI Case of an evenly decreasing load on l, i.e. a = 0 and pl = 0: p l2 5 pal 3 MMAX = MA = a θMAX = θB = yMIN = yA = 24 EI 3


Straight beams

− 5 pl 4 24 EI − 3 pl l 4 40 EI − 2 pa l 4 15 EI

Issue 1

SSM V2-4 • BENDING OF BEAMS 2 - DISTRIBUTED LOAD COMPLEMENT: case of loading non contiguous to B The case of a load not extended to point B may be processed by applying the principle of superposition. pl pb pa pa pa a a b

pb

pl

b A

B

A

B

(1)

A

B

(2)

End values The end value equations are written as the sum of the equations giving the values in each case (1) and (2), i.e.: RTotal A = R (A1) + R (A2) MTA = M (A1) + M (A2) θTA = θ (A1) + θ (A2) yTA = y (A1) + y (A2)

θTB = θ (B1) + θ (B2)

RTB = R (B1) + R (B2)

MTB = M (B1) + M (B2) l−a by writing: pl = pa + (pb - pa) b−a

yTB = y (B1) + y (B2)

General expressions With the end values known, it is possible to calculate the T, M, θo, and ∆o values at any point along the span using the general equations below:

[

]

pb − pa 2 2 x−a − x−b , 2(b − a ) p p p − pa 3 MT = MTA + RTAx - a áx - añ2 + b áx - bñ2 - b x−a − x−b 2 2 6 (b − a)

TTotal = RTA - pa áx - añ + pb áx - bñ -

[

θT = θTA +

p M TA x RTA x 2 + − a x−a EI 2EI 6 EI

yT = yTA + θTAx +

3

+

pb x−b 6 EI

MTA x 2 RTA x 3 p + − a x−a 2EI 6 EI 24 EI


4

+

3

], p −p − x−a 24 EI (b −a ) [ b

pb x−b 24 EI

a

4

Straight beams

−

3

4

− x−b

pb − pa 120 EI ( b −a )

[ x−a

4

5

], − x−b

5

].

page V2-4•2-9

SSM V2-4 • BENDING OF BEAMS 3 - LOADING MOMENT

y

General expressions Shear force: T = RA Mo Bending moment: M = MA + RAx + Mo áx - añ0 θB M x R x2 M x yA Slope: θ = θA + A + A + o áx - añ MA EI EI 2EI RB 2 θA M x R x3 M Deflection: y = yA + θAx + A + A + o áx - añ2 l RA 2EI 6 EI 2EI If x < a thenáx - añ0 = 0 ; if x > a then áx - añ0 = 1 ; by extension,áx - añn is interpreted as the product (x - a)n áx - añ0. end on a single end blocked in clamped end free end rotation bearing point

a

MB

3-a a

Mo

A

RA = 0

MA = 0

RB = 0

MB = Mo

3-b a

A

RA = 0

Mo

RB = 0

B

MMAX = MB ; if a = l, MMAX = Mo

Remarkable values Mol 2 ; ∀x, y > 0 yMAX = yA ; if a = 0, yMAX = 2EI

l−a MA = - Mo l a MB = Mo l

MMIN = MA ; if a = 0, MMIN = - Mo

3-c a

A

End values Mo 2 2 (l - a ) yA = 2EI yB = 0

B − Mol θMAX = θA ; if a = 0, θMAX = EI

MMAX = Mo

− Mo θA = (l - a) EI θB = 0

Mo

B

− 3 Mo l 2 − a 2 RA = 2 l3 3 Mo l 2 − a 2 RB = 2 l3

θA = 0

End values Mo a(l - a) yA = 2EI

θB = 0

yB = 0

Remarkable values l M l2 yMAX = yA ; if a = , yMAX = o ; ∀x, y > 0 2 8 EI

End values − Mo (l − a ) (l − 3a ) MA = 0 θA = yA = 0 l 4 EI − Mo l 2 − 3a 2 MB = θB = 0 yB = 0 l2 2 Remarkable values

ì Mo 2l 3 − 3l 2 + 3a 3 for x = a + ε ; if a = 0 or a = l, M MAX = Mo ï M MAX = l3 2 ï ì ï − M o l 2 − 3a 2 − Mo M í for x = l ; if a = 0, M MIN = ïï if a < 0.282 l: M MIN = 2 2 2 l ï M MIN í 2 2 ï ï if a > 0.282 l: M MIN = − 3 Mo a(l − a ) for x = a − ε ; if a = 0.577 l, M MIN = − 0.577 Mo ïî ïî l3 2

l ì ì ï ïï If a ≤ 3 the case y > 0 does not exist. ï y MAX í M l2 M (l − a) ( 3a − l ) 3/ 2 3a − l ï ; if a = 0.721 l (therefore x = 0.475 l ), y MAX = 2.57 x 10 −2 o for x = l ï y MAX = o ï 6 EI 3(l + a) EI 3(l + a) ïî ï ì ïï ï If a ≥ 0.577 l the case y < 0 does not exist. y í ï ï 2 4 ù é ïï − Mo (l 2 − a 2 ) x 3 − 2l 3 ( x − a) 2 + l 2 (l − 3a) (l − a) x ï 2l 3 1 æ aö æ aö ê1 − 1 − 6ç ÷ + 9ç ÷ ú ; for x = ï y MIN í y MIN = 4 EI 3 2 2 èlø èlø ú 2 3(l − a ) ê l ï ï ë û ï ï − Mo l 2 l ï ï if a =0 (therefore x = ), y MIN = ïî ïî 3 27 EI


Straight beams

Issue 1

SSM V2-4 • BENDING OF BEAMS 3 - POINT MOMENT

3-d

End values a( l − a ) RA = - 6 Mo l3

a Mo

A

B

RB = - RA

l 2 − 4al + 3a 2 MA = - Mo l2 2al − 3a 2 MB = - Mo l2

θA = 0

yA = 0

θB = 0

yB = 0

Remarkable values ì 4al − 9 a l + 6 a for x = a + ε ; if a = l, MMAX = Mo ïï MMAX = Mo l3 M í 3 2 2 3 ï MMIN = − Mo l − 4al + 9 a l − 6 a for x = a − ε ; if a = 0, MMIN = − Mo ïî l3 2

2

3

l ì ì ï ïï If a ≤ 3 , the case y > 0 does not exist. ï y MAX í − M o (l 2 − 4al + 3a 2 ) 3 M l2 l( 3a − l ) ï ï y MAX = for x = if a = 0.768 l ( therefore x = 0.566 l ), y MAX = 1.615 x 10 −2 o 2 2 ïï ïî 54 EI 3a EI a (l − a ) y í 2l ì ï ïï If a ≥ 3 , the case y < 0 does not exist. ï y í MIN ï − M o ( 2al − 3a 2 ) 3 M l2 l2 ï y MIN = ; if a = 0.232 l ( therefore x = 0.434 l ), y MIN = − 1.615 x 10 −2 o for x = ï 2 2 54 EI a (l − a) 3(l − a) EI ïî ïî

3-e

End values

a

Mo

A

B

− Mo RA = l

MA = 0

R B = - RA

MB = 0

− Mo 2l 2 − 6 al + 3a 2 θA = 6 EI l Mo l 2 − 3a 2 θB = 6 EI l

yA = 0 yB = 0

Remarkable values l−a ì ï MMAX = Mo l for x = a + ε ; if a = 0, MMAX = Mo M í a ï MMIN = − Mo for x = a − ε ; if a = l, MMIN = − Mo l î

ì ì If a ≤ 0.423 l, the case y > 0 does not exist. ï ï 2 Mo (− 2l 2 + 6al − 3a 2 ) 3/ 2 2l 2 2 −2 Mo l ï y MAX í y x = for = − + 2 − ; = ( therefore = 0 . 577 ), = 6 . 42 10 x al a if a l x l y MAX ï MAX 9 3EI ï 3 l EI î y í If a ≥ 0.577 l, the case y < 0 does not exist. ì ï ï 2 ï y MIN í − Mo (l 2 − 3a 2 ) 3/2 l2 2 −2 Mo l = x = − − ; if = 0 ( therefore = 0 . 423 ), = − 6 . 42 10 y for x l a a x l y ï MIN MIN ï 3 l EI 9 3EI î î

3-f

a

A

Mo

B

RA = 0

MA = - Mo

θA = 0

RB = 0

MB = 0

θB =

− Mo a EI

End values Mo a( 2l − a ) yA = 2EI yB = 0 Remarkable values

MMAX = Mo for 0 ≤ x < a

− Mol EI Mol 2 ; ∀x, y > 0 yMAX = yA ; if a = l, yMAX = 2EI

θMIN = θB ; if a = l, θMIN =


Straight beams

page V2-4•2-11

SSM V2-4 • BENDING OF BEAMS 4 - IMPOSED DEFLECTION

∆o

y

General expressions

Shear force: T = RA - α∆oEI áx - añ Bending moment: M = MA + RAx - α∆oEI áx - añ x ∆ M x R x2 2 yA MA Slope: θ = θA + A + A − α o x − a RB EI 2EI 2 θA ∆ M x2 R x3 3 l Deflection: y = yA + θAx + A + A − α o x − a RA 2EI 6 EI 6 If x < a then áx - añ0 = 0 ; if x > a then áx - añ0 = 1 ; by extension,áx - añn is interpreted as the product (x - a)n áx - añ0. end on a single end blocked in clamped end free end rotation bearing point 0

MB

a

θB

3 (l − a ) 3

α=

4-a

End values

a

RA = 0

∆o

A

B

RB = 3∆o

α=

4-b a

∆o

MMAX = MA ; if a = 0, MMAX = 6∆o

B

EI l2

yA =

− ∆ o 2l − 3l a + a 3 2 (l − a ) 3 2

θB = 0


yMIN = yA ; if a = 0, yMIN = - ∆o

12l (l − a ) (l + 2al − 3a 2 ) 2

2

End values l a(l + 2a ) MA = 6∆oEI 2 θA = 0 yA = ∆o 2 RA = 0 l + 2al − 3a 2 l + 2al − 3a 2 12∆ o EIl − 6 ∆ o EI l+a RB = MB = 2 θB = 0 yB = 0 (l − a ) 2 (l 2 + 2al − 3a 2 ) l + 2al − 3a 2 l − a Remarkable values EI MMIN = MB ; if a = 0, MMIN = - 6∆o 2 yMIN = yA ; if a = 0, yMIN = - ∆o l

α=

4-c a

A

EI (l − a ) 3

3∆ o 2(l − a ) EI MB = 3∆o (l − a ) 2

θA =

3∆ o θMAX = θA ; if a = 0, θMAX = 2l

EI MMIN = MB ; if a = 0, MMIN = 3∆o 2 l

A

MA = 0

3

12l 3 a 2 (l − a ) 2 ( 3l 2 − 2al − a 2 ) End values

∆o

B

RA = 6∆oEI RB =

2l + a a 2 ( 3l 2 − 2al − a 2 )

6 ∆ o EI ( 3l 2 − a 2 ) a(l − a ) ( 3l − 2al − a ) 2

2

2

MA = 0 MB =

θA = - 3∆o

2

l a( 3l 2 − 2al − a 2 )

− 6 ∆ o EIl (l 2 − a 2 ) a(l − a ) 2 ( 3l 2 − 2al − a 2 )

yA = 0

θB = 0 yB = 0 Remarkable values

ì ï MMAX

M í

ïî M MIN

2l + a EI EI = 6∆o for x = a ; if a = 0.366 l, M MAX = 18.1753∆ o 2 2 2 ( 3l − 2al − a ) a l = M B ; if a = 0.5773 l, MMIN = − 25.6474∆ o EIl −2

1/ 2 1/ 2 ì æ a ö æ a ö l3 ï if a > 0.414 l: y MIN = − 2∆ o for x = l ç ç ÷ ÷ a( 3l 2 − 2al − a 2 ) è ( 2l + a ) ø è 2l + a ø ïï y í if a =0 .414 l :y MIN = − ∆ o for x = a ï l3 (l − a) (l + a) 3 (l 2 + a 2 ) for x = l 2 ï if a < 0.414 l: y MIN = − 4 ∆ o 2 2 2 2 2 a( 3l − 2al − a ) ( 3l − a ) 3l − a 2 ïî


Straight beams

Issue 1

SSM V2-4 • BENDING OF BEAMS 4 - IMPOSED DEFLECTION

α=

4-d a

3l 3 a (l − a) 3 3

End values B

A

∆o

l + 2a RA = 3∆oEI 3 a (l − a ) 3l − 2a RB = 3∆oEI a (l − a ) 3

MA = - 3∆oEI

l

a 2 (l − a ) l MB = - 3∆oEI a (l − a ) 2

θA = 0

yA = 0

θB = 0


EI ìM ï MAX = 6 ∆ o a(l − a ) for x = a ; if a = M í l l ï if a < : M MIN = M A ; if a = , M MIN 2 3 î yMIN

l EI , M MAX = 24∆ o 2 2 l EI 81 =− ∆o 2 2 l

ì l l3 2al for x = ïï if a < : yMIN = − 2∆ o 2 (l + 2a ) (l − a ) 2 1 + 2a í ï if a = l : yMIN = − ∆ o for x = a ïî 2

α=

4-e

3l a 2 (l − a ) 2

a

End values l a 2 (l − a ) l RB = 3∆oEI a (l − a ) 2

RA = 3∆oEI A

∆o

B

MA = 0 MB = 0

2l − a 2a(l − a ) l+a θB = ∆o 2a(l − a )

θA = - ∆o

yA = 0 yB = 0 Remarkable values

l EI EI MMAX = 3∆o for x = a ; if a = , MMAX = 12∆o 2 2 l a( l − a )

yMIN

3/ 2 1/ 2 ì æ l 2 − a2 ö l − ∆ o æ l 2 − a2 ö ï if a < : yMIN = x l for = − ç ÷ ç ÷ ï 2 a (l − a ) 2 è 3 ø è 3 ø í ï if a = l : y MIN = − ∆ o for x = a ïî 2

α=

4-f

3 (l − a ) 2 (l + 2a )

a

A

End values

∆o

B

1 RA = 3∆oEI 2 a (l − a ) 1 RB = 3∆oEI a (l − a ) 2

MA = 0 MB = 0

2l − a θA = - ∆o 2a(l − a ) l+a θB = ∆o 2a(l − a )

yA = 0 yB = 0

Remarkable values

MMAX = MA for 0 ≤ x < a ; ifa = 0, MMAX = 3∆o 3 ∆o 2 l yMIN = yA ; ifa = 0, yMIN = - ∆o

EI l2

θMAX = θB ; ifa = 0, θMAX =


Straight beams

page V2-4•2-13

SSM V2-4 • BENDING OF BEAMS 5 - IMPOSED ROTATION

θo

y

General expressions

MB

Shear force: T = RA Bending moment: M = MA + RAx + αθoEI áx - añ0 x M x R x2 yA MA Slope: θ = θA + A + A + αθo áx - añ RB EI 2EI θA θ M x2 R x3 l Deflection: y = yA + θAx + A + A + α o áx - añ2 RA 2EI 6 EI 2 If x < a then áx - añ0 = 0 ; if x > a then áx - añ0 = 1 ; by extension,áx - añn is interpreted as the product (x - a)n áx - añ0.

a

θB


free end

α=

5-a θo


clamped end

1 l−a

End values

a

A

B

RA = 0

MA = 0

RB = 0

MB = θo

EI l−a

θA = - θo

l+a yA = θo 2

θB = 0


EI MMAX = θo for a < x ≤ l l−a θMAX = θo for 0 ≤ x ≤ a

yMAX = yA ; ifa = 0, yMAX = θo

l ; ∀x, y > 0 2 α=

5-b θo

l a( l − a )

End values

a RA = 0 A

B

RB = 0

EI a EI MB = θo l−a MA = - θo

θA = 0

yA = θo

θB = 0

yB = 0

l 2

Remarkable values

MMAX = MB ; ifa → l, MMAX → + ∞ MMIN = MA ; ifa → 0, MMIN → - ∞ yMAX = yA ; ∀x, y > 0


Straight beams

Issue 1

SSM V2-4 • BENDING OF BEAMS 5 - IMPOSED ROTATION

α=

5-c θo

[

4l 3

(l − a ) (l − a ) 3 + 4 a 3

] End values

a RA = - 6θoEI A

B

R B = - RA

1+ a ( 1 − a ) 3 + 4a 3

θA = - θo

MA = 0

MB =- 2θo EI

l ( l − 3a ) ( l − a ) 3 + 4a 3

yA = 0

θB = 0

yB = 0

l ( l 2 − 3a 2 )

[

(l − a ) (l − a ) 3 + 4 a 3

2

]

Remarkable values ì EI 2l − 3al + 3a EI if a → l, M MAX → + ∞ for x = a + ε ; if a = 0, M MAX = 4 θ o ï M MAX = 2 θ o 3 3 l a l − l a a ( ) − + 4 ï ï EI ì M í if a < 0.282 l: M MIN = M B ; if a = 0, M MIN = − 2 θ o l ï M MIN ïí EI ï ïif a > 0.282 l: M MIN = R A a for x = a − ε ; if a = 0.577 l, M MIN = − 6.468 θ o ïî l î 3

2

3

l ì ì ï ïï If a ≤ 3 the case y > 0 does not exist. ï y MAX í 3a − l ( 3a − l ) 3 / 2 2 l3 ï for x = l ; if a = 0.677 l ( therefore x = 0.453 l ), y MAX = 0.244 θ o l ï y MAX = θ o 3 3 ï 3 3(l + a ) ( l − a ) + 4a 3(l + a) ïî ï ì ï y í ï If a ≥ 0.577 l the case y < 0 does not exist. ï ï 2 4 ù é ù é −θo 2l 3 2l 3 ï 1 ï æ aö æ aö ê1 − 1 − 6ç ÷ + 9ç ÷ ú ; (l + a ) x 3 − ( x − a) 2 + l 2 (l − 3a ) x ú for x = y ï MIN í y MIN = 3 3 ê 2 2 è lø ú èlø 1− a 2 3(l − a ) ê ( l − a ) + 4a ë û ï ï ë û ï ï −4 l = = = − = = = a l x l y l a x y l θ θ if 0 . 219 ( therefore 0 . 4 ), 0 . 253 ; if 0 ( therefore ), ï ï MIN o MIN o 3 27 î î

5-d

α=

θo

l3 a(l − a ) (l 2 − 3al + 3a 2 ) End values

a RA = - 6θo A

B

R B = - RA

EI l 2 − 3al + 3a 2

l(l − 3a ) a(l 2 − 3al + 3a 2 ) l ( 2l − 3a ) MB = - θoEI (l − a ) (l 2 − 3al + 3a 2 )

MA = - θoEI

θA = 0

yA = 0

θB = 0

yB = 0

Remarkable values ì 4al − 9 a l + 6 a EI for x = a + ε ; if a → l, M MAX → + ∞ ï MMAX = θ o ï a (l − a ) (l 2 − 3al + 3a 2 ) M í 3 2 2 3 ï MMIN = − θ o EI l − ( 4al − 9 a l + 6 a ) for x = a − ε ; if a → 0, MMIN → − ∞ 2 2 ïî a (l − a ) (l − 3al + 3a ) 2

2

3

ì ì If a ≤ l , the case y > 0 does not exist. ï ïï 3 ï y MAX í 3 −θo æ l ö ( l − 3a ) 3 l ( 3a − l ) ï ï y MAX = ; if a = 0.735 l ( therefore x = 0.547 l ), y MAX = 0.1963 θ o l for x = ç ÷ 2 ïï ïî 54 è a ø l − 3al + 3a 2 3a y í 2l ì ï ïï If a ≥ 3 , the case y < 0 does not exist. ï 3 ï y MIN í θ æ l ö ( 3a − 2l ) 3 l2 ï y MIN = o ç ; if a = 0.265 l ( therefore x = 0.453 l ), y MIN = − 0.1963 θ o l ÷ 2 ï 2 for x = 54 è l − a ø l − 3al + 3a 3(l − a ) ïî ïî


Straight beams

page V2-4•2-15

SSM V2-4 • BENDING OF BEAMS 5 - IMPOSED ROTATION CONTINUED y

MB

a yA

General equations

θo θB x

MA

RB

θA l

RA

Shear force: T = RA Bending moment: M = MA + RAx + αθoEI áx - añ0 M x R x2 Slope: θ = θA + A + A + αθo áx - añ EI 2EI θ M x2 R x3 Deflection: y = yA + θAx + A + A + α o áx - añ2 2EI 6 EI 2

If x < a then áx - añ0 = 0 ; if x > a then áx - añ0 = 1 ; by extension,áx - añn is interpreted as the product (x - a)n áx - añ0.


free end

5-e

α=

θo


clamped end

3l l − 3al + 3a 2 2

End values

a

A

B

EI RA = - 3θo 2 l − 3al + 3a 2

MA = 0

− θ o 2l 2 − 6 al + 3a 2 θA = 2 l 2 − 3al + 3a 2

RB = - RA

MB = 0

θB =

θo l 2 − 3a 2 2 2 l − 3al + 3a 2

yA = 0 yB = 0

Remarkable values l−a EI ì ïï MMAX = 3θ o EI l 2 − 3al + 3a 2 for x = a + ε ; if a = 0, M MAX = 3θ o l M í a EI ï MMIN = − 3θ o EI 2 for x = a − ε ; if a = l, MMIN = − 3θ o 2 ïî l − 3al + 3a l

ì ì ï ï If a ≤ 0.423 l, the case y > 0 does not exist. ï ï θ ( − 2l 2 + 6 al − 3a 2 ) 3/ 2 1 ï ïy y MAX = o for x = ( − 2l 2 + 6 al − 3a 2 ) ; if a = 0.718 ( therefore x = 0.5 l ), í MAX ï 3 l 2 − 3al + 3a 2 3 3 ï ï θ l ï ï y MAX = 0.3257 θ o l ; if a = l ( therefore x = 0.577 l ), y MAX = o ïî ï 3 3 í ì ï ï If a ≥ 0.577 l, the case y < 0 does not exist. ï ï ï θ o ( l 2 − 3a 2 ) 3 / 2 1 2 ï =l− (l − 3a 2 ) ; if a = 0.282 l ( therefore x = 0.5 l ), y MIN = − 0.3257 θ o l ; ï y MIN í y MIN = 2 2 for x 3 − + 3 3 l al a 3 3 ï ï − θ ol ï ï if a = 0 ( therefore x = 0.423 l ), y MIN = ï ïî 3 3 î

y

5-f

α=

θo

1 a End values

a

A

B

RA = 0

MA = - θo

RB = 0

MB = 0

EI a

θA = 0 θB = - θo

θo (2l - a) 2 yB = 0 Remarkable values

yA =

EI for 0 ≤ x ≤ a a θMIN = θo for a ≤ x ≤ l

MMAX = θo

yMAX = yA ; ifa = l, yMAX = θo

l ;∀x, y > 0 2


Straight beams

Issue 1

SSM V2-4 • BENDING OF BEAMS V2-4 2.2 EXAMPLES EXAMPLE 1 y

pa pb

a

EI = 84.108 Nmm² pa = 20 N/mm pa = 5 N/mm l = 800 mm a = 210 mm

x b

b = 600 mm

l A

B

CALCULATION OF END VALUES Table 2-c gives the equations to be used in the case of loading contiguous to B (i.e.b=l):

pa (l − a) 3 ( 3l + a) pl − pa (l − a) 3 ( 4l + a) − pa (l − a) 3 (l + 3a) pl − pa (l − a) 3 ( 2l + 3a) + θ = , − A 48 EI l 240 EI l 8 40 l3 l3 p + pl 2 pa + pl RB = - RA + a (l - a) MB = RAl (l - a)2 ; 2 6 and yA = 0 MA = 0 θB = 0 yB = 0.

RA =

The application of the principle of superposition (refer to table 2-"complement") gives, by writing l−a , pl = pa + (pb - pa) b−a with, in case (1)

pa(1) = pa = 20 N/mm,

pl( 1) = pl =

1 N/mm, 3

and in case (2)

pa(2) = - pb = - 5 N/mm,

pl( 2) = - pl = -

1 N/mm, 3

at RTotal A = RA(1) + RA(2) = 1944.85 + (- 7.84) = 1937 N θTA = θ (A1) + θ (A2) = - 1.263 x 10-2 + 8.98x 10-5 = - 1.25 x 10-2 rd (- 0.718°)

RTB = RB(1) + RB(2) = 4035.48 + (- 365.49) = 3688 N MTB = M B(1) + M B(2) = - 784.1 x 103 + 27.5x 103 = - 756.6 x 103 mmN

RTB may also be calculated using the general equations in table 2-"complement", i.e.: RTB = - TT (x = l),

{x = l} Þ

{ x−a

RTB = - RTA + pa (800 - a) - pb (800 - a) +

n

= ( x − a) n ; x − b

n

= ( x − b)n

}

pb − pa ( 800 − a) 2 − ( 800 − a) 2 2( b − a )

[

]

= - 3688 N. The sign convention of paragraph 2.1.2 indicates that RTA and RTB are directed upwards, MTB compresses the fibres at y < 0 and that θTA is oriented clockwise.


Straight beams

page V2-4•2-17

SSM V2-4 • BENDING OF BEAMS CALCULATION OF MTMAX x is calculated such that TT = 0, i.e. TT = RTA - pa áx - añ + pb áx - añ -

{

- for x ≤ a ; x − a

}

n

= 0 ;x − b

n

= ( x − a) n ; x − b

n

p b − pa

x−a 2( b − a) [

2

− x−b

2

] = 0:

= 0

e

TT = ct ;

{

- for x ≥ b ; x − a

n

= ( x − b)n

}

p − pa dTT [2(x - a) - 2(x - b)] = 0 therefore TT = cte ; = - pa + pb - b dx 2( b − a )

{

- for a ≤ x ≤ b ; x − a

n

TT = RTA - pa(x - a) -

= ( x − a)n ; x − b

n

}

=0

pb − pa (x - a)2 ; {TT = 0} ⇔ {x = 316,3 mm} ; 2( b − a )

the equation of the bending moment being p p p − pa MT = MTA + RTAx - a áx - añ2 + b áx - bñ2 - b 2 2 6 (b − a ) the value sought is

MTMAX

= MT (x = 316.3) = MTA + 316.3 RTA -

[ x−a

3

− x−b

3

],

pa p − pa (316.3 - a)3 (316.3 - a)2 - b 2 2( b − a )

= 506.3 x 103 mmN ; MTMAX compresses the fibres at y > 0.

CALCULATION FOR ANY VALUE OF x

θT1, θT2, θT3 are calculated respectively for x = 100 mm, x = 400 mm and x = 700 mm. M x R x2 p p pb − pa 3 3 4 x−a − x−b θT = θTA + TA + TA − a x − a + b x − b − EI 2EI 6 EI 6 EI 24 EI (b − a )

[

{

For x = 100 ; x − a

n

= 0 ;x − b

n

4

]

}

= 0 ,

MTA x RTA 100 2 + 2EI EI = - 1.14 x 10-2 rd (- 0.625°) ; θT1 is oriented clockwise.

θT1 = θTA +

{

For x = 400 ; x − a

n

= ( x − a)n ; x − b

n

}

=0 ,

MTA x RTA 400 2 p pb − pa + − a ( 400 − a) 3 − (400 - a)4 EI 2EI 6 EI 24 EI (b − a) = 0.34 x 10-2 rd (0.195°) ; θT2 is oriented counter clockwise.

θT2 = θTA +

{

For x = 700 ; x − a

n

= ( x − a) n ; x − b

n

}

= ( x − b)n ,

MTA x RTA 700 p p pb − pa + − a (700 − a) 3 + b (700 − a) 3 − (700 − a) 4 − (700 − b) 4 EI 2EI 6 EI 6 EI 24 EI ( b − a) = 0.68 x 10-2 rd (0.390°) ; θT3 is oriented counter clockwise.

θT3 = θTA +


[

2

Straight beams

]

Issue 1

SSM V2-4 • BENDING OF BEAMS GRAPHIC ILLUSTRATION y

Equilibrium

3688 N

20 N/mm 5 N/mm

1937 N

- 756.6 x 103 mmN

x A 1937

2000

B

1000

x (mm)

x = 316.3

0

0

100

200

300

400

500

600

700

800

-1000

TT (N)

-2000 -3000 -4000

x=a

-3688

Shear force

[

TT = 1937 - 20áx - añ + 5áx - bñ + 1.667 x 10-2 x − a 600

2

− x−b

2

]

x=b

506.3

400

x = 316.3

200

x (mm) 0

100

200

MT (103 mmN)

0 -200 -400 -600

300

x=a

-800

400

500

Bending moment

MT = - 756.6 x 10 + 1937x - 10áx - añ + 2.5áx - bñ + 5.556 x 10 3

0

θ = -1.25 x 10-2

600

2

200

300

2

400

0

500

-3

[ x−a

700

800

x=b 3

− x−b

3

600

]

-756.6

700

x (mm)

800

θ = 0.68 x 10 rd -2

-0.5

θ = -1.14 x 10-2 rd

-1

x = 700

-1.5

x = 100

-2

θ = 0.34 x 10-2 rd x = 400

-2.5

-3.5

x=a

yT (mm)

-3

x=b Deflected shape

[

EI yT = - 105.34 x 10 x - 378.3 x 10 x + 322.8 x - 0.833áx - añ4 + 0.208áx - bñ4 + 2.778 x 10-4 x − a 6

3

2

3

[

EI θT = - 105.34 x 106 - 756.6 x 103 x + 1937 x2 - 3.333 áx - añ3 + 0.833 áx - bñ3 + 1.389 x 10-3 x − a Issue 1 © AEROSPATIALE - 1999

Straight beams

5 4

− x−b − x−b

] ]

5 4

page V2-4•2-19

SSM V2-4 • BENDING OF BEAMS EXAMPLE 2

P C I

J

EI = 84 x 108 Nmm2 P = 5000 N C = - 1 x 106 mmN d = 800 mm d1 = 100 mm d2 = 500 mm

K

d1 d2 d

CALCULATION OF REACTIONS Transfer of force P from I to J and application of the principle of superposition as schematised.

P

I

C

J

y

K

y

y

F=P

Mo = - d1P

Mo = C

x A

x

x

B

A

a = 0 l = d - d1 (1)

B

A B a = d 1 - d 2 l = d - d1 (3)

a = 0 l = d - d1 (2)

giving for span [J, K] at: RJ = RA(1) + RA(2) + RA(3) = 5000 + 1071.43 + 1443.15 = 7515 N RK = RB(1) + RB(2) + RB(3) = 0 - 1071.43 - 1443.15 = - 2515 N MK = M B(1) + M B(2) + M B(3) = 0 + 250 000 + 10 204 = 260.2 x 103 mmN

ì (1) F (l − a) 2 ( 2l + a) ï RA = 2 l3 ï 2 F a( 3l − a 2 ) ï where í RB(1) = 2 l3 ï 2 2 ï M (1) = − F a(l − a ) B 2 ï 2 l î (table 1-c)

RA(2)(3) =

− 3 Mo l 2 − a 2 2 l3

RB(2)(3) = − RA(2)(3) M B(2)(3) =

− M o l 2 − 3a 2 2 l2

(table 3-c)

CALCULATION OF THE DEFLECTED SHAPE AT POINT I Considering spans [I, J] and J, [ K] mutually fixed atJ by a continuation of the part, the deflection at I is the algebraic sum of: - the deflection due to P if the beam, reduced in span [I, J] was fixed atJ, - and the deflection caused by rotation of the section at J.

y

y P

θJ

x I

J

I

J

(4)


x K

(5) Straight beams

Issue 1

SSM V2-4 • BENDING OF BEAMS Therefore, we obtain: (5) yI = y (4) I + y I and

−F (2l3 - 3l2a + a3) ; table 1-a, whereF = P, l = d1, a = 0 ; 6 EI = - 0,1984 mm = xI θJ = - d1 θJ.

y I(4) = y I(5)

The same calculations give, in the case of rotation at: (5) θI = θ (4) I + θ I and

F (l - a)2 ; table 1-a, whereF = P, l = d1, a = 0 ; 2EI = 2.976 x 103 rd = θJ.

θ (4) = I θ (5) I

The assumptions taken for the calculation of reactions (refer to previous page) entail for the determination of θJ at:

− M o ( l − a ) (l − 3 a ) ; table 3-c, with case (2):Mo = - d1F, l = d - d1, a = 0 ; 4 EI l case (3): Mo = C, l = d - d1, a = d2 - d1 ; θJ = 10.417 x 10-3 + (- 6.378 x 10-3) = 4.039 x 10-3 rd (0.231°). (3) θJ= θ (2) A + θ A and

θ (2)(3) = A

The deflected shape at I is finally characterised by:

yI = - 0.1984 + (- 100) x 4.039x 10-3 = - 0.602 mm

θI = 2.976x 10-3 + 4.039x 10-3 = 7.015x 10-3 rd (0.402°).

and

GENERAL EXPRESSIONS Whatever the type of load, the general expressions are similar. Each equation includes the components due to boundary conditions at A (end at x = 0) and the component expressing the loading effect:

y due to concentrated load: üï ý y due to applied moment: ïþ

F 3 ì M A x 2 RA x 3 ï − 6 EI x − a yA + θ A x + + í 2EI 6 EI ï + Mo x − a 2 î 2EI boundary condition components/loading component

For example, writing the deflection equation boils down to summing the effects of the various components, i.e. in this case, by assimilating the reaction at J to a supplementary loading:

y

y

P

PI

C

PJ C x

x I

J

K

I

J

K

By writing PJ = -RJ (due to sign convention, refer to paragraph V1-4 2.1.2) and PI = P. Boundary condition components: yI ≠ 0 ;θI ≠ 0 ; M I → 0 ; R I → 0. Loading components: P P x3 3 PI → − I x − 0 = − I 6 EI 6 EI


; PJ → −

PJ x − d1 6 EI

3

; C→

Straight beams

C x − d2 2EI

2

.

page V2-4•2-21

SSM V2-4 • BENDING OF BEAMS The deflection expression for any x is then:

y = yI + θIx -

P PI x 3 − J x − d1 6 EI 6 EI

3

or

y = yI + θIx +

C x − d2 2EI

2

−

+

C x − d2 2EI

[

2

,

1 Px 3 − RJ x − d1 6 EI

By successive derivations, we obtain: C 1 θ = θI + x − d2 − Px 2 − RJ x − d1 EI 2EI M = Cáx - d2ñ0 - Px + RJáx - d1ñ T = - P + RJáx - d1ñ0

[

2

3

]

(1).

]

(2), (3), (4).

These expressions make up a system of four equations with three unknowns (yI, θI and RJ) which are evaluated by solving the system {θ(x = d) = 0 ; y(x = d) = 0 ; y(x = d1) = 0}. Once the specific values at I and the reaction at J have been established, these equations also make it possible to calculate T, M, θ and y for any x. For example, for x = d1, the θ equation gives 4.039 x 10-3 and verifies the calculation in the previous paragraph. The reactions at K are determined by M(x = d) = MK and T(x = d) = - RK (refer to sign convention, paragraph V1-4 2.1.2).

GRAPHIC ILLUSTRATIONS

y (mm)

θ (10-3 rd)

0,75 0,5 0,25 0

7.5 5 2.5

x (mm)

0

100

200

300

400

500

600

0 800

700

-2.5

-0,25

-5

-0,5 -0,75

-7.5

Deflected shape

5000

M (103 mmN)

600 400

3000

200 1000 0

0

100

200

300

400

500

600

700

x (mm)

800 -1000

-200 -3000

-600


T (N)

-400

Bending moment and shear force

Straight beams

-5000

Issue 1

SSM V2-4 • BEAM BENDING V2-4

3 HIGH CURVATURE BEAMS

V2-4

3.1

GENERAL INFORMATION

The qualification "curved beam" is adopted as soon as an initial curvature radius exists - radius included in the bending plane. This geometrical characteristic entails: - a bending mode which is always the composed bending mode (resulting T, M and N for a straight section), - a specific distribution for normal stresses. The purpose of this chapter is to provide practical calculation information to quantify the effect of the initial curvature on stress values. In practice, this effect can only be assessed by evaluating the R/h ratio in which R is the initial curvature radius of the average line and h the height of the section.

R ≤ 5: case of high curvatures, special formulas. h R > 5: case of arcs, usual formulas for in-plane bending. h R = 5: error of around 7% (*). h R = 10: error of around 3.5% (*). h R = 15: error of around 2% (*). h (*) given for information purposes for a rectangular section.

V2-4

3.2

BASIC THEORY

The assumptions are those of in-plane bending, indicated in paragraph 2.1.1. The main difference in relation to straight beams resides in the difference of the initial length of fibres in function of the radius, entailing: - a non-linear distribution of normal stresses - the stress remains proportional to the deformation but is not proportional to the distance to the neutral axis, - a neutral axis that does not pass by the centre of the section. a) single bending

e

σ

σ G

y

G

A.N.

A.N. r

(S)

M

M rG

rN 0'

0


High curvature beams

0

RG : rN: r: G: A.N.:

initial curvature radius neutral fibre radius radius of a calculation point centre of the section neutral axis

page V2-4•3-1

SSM V2-4 • BEAM BENDING Hooke's law, then sectionS() equilibrium, lead to a normal stress of the following form: Hooke's law → σx = E εx = E ω

y , rN + y

ì σ dS = N = 0 x ï ï( S ) equilibrium → í ï σ x ydS = M ïî(S)

ò

ò

Þ

M ì ïω = ESe , ï S írN = . dS ï ï r (S) î

ò

This stress obeys a hyperbolic distribution and increases very quickly at internal fibres whenR/h decreases. The practical equation is: y M σx = . Se rN + y b) normal load In this case, the equilibrium in the section gives: ì σ dS = N x ï N ìï ï( S ) ω= , Þ í í ES ïîrN = 0. ï σ x ydS = M = 0 ïî( S )

ò ò

Therefore there is no effect of initial curvature on the contribution of the normal load in the calculation of σx. c) combined bending By superposition, the stress becomes: N M y σx = + . S S( rG − rN ) rN + y d) shear load For curved beams under shear and bending, the negligible effect of curvature on longitudinal shear stress is proven:

τexact = k x τ(neglected curvature) with 1 < k < 1.1. Therefore, τ is calculated with straight beam formulas. V2-4

3.3

PRACTICAL FORMULAS

The position of the neutral axis given by rN is the only unknown factor in the σx equation. The following table calculates the normal stress value for most cross sections encountered using four distinct formulas. For tubular sections, the difference between rN(total cross section) and rN(empty cross section) must be determined. For the case of any type of cross section it is therefore necessary to calculate the integral

ò dS / r

(reducing dS to

the x dr form) or, if this proves to be difficult, to break down the cross section into simple elements to approach the rN value (refer to composed cross sections below). It is important to note that rN is independent from the width b of the cross section (refer to "rectangular cross section" and "elliptic cross section"); for combined sections, rN only depends on the width ratios between the sections (refer to trapezoidal sections and combined sections). For example, this makes it possible to assimilate a parallelogram, the base of which perpendicular to r, with a rectangle of the same height. This is only valid for the calculation of rN, S has to be calculated with the characteristics of the real cross section.



Revision 0

SSM V2-4 • BEAM BENDING M y S(rG − rN ) rN + y

Normal bending stress:

Rectangular cross section

rN =

h

S bh h = r → rN = e dS r dr ln e b r ri (S) r

ò

ò ri

re

Square cross section, side a h

ri

make h = a

b

0

0' be

Trapezoidal cross section h( bi + be )

rN =

é b r − be ri re ù 2ê i e ln − ( bi − be )ú h ri ë û be h(1 + α ) where rN = with α = bi é re − αri re ù 2ê ln − (1 − α )ú ri ë h û

h re ri

bi

Triangular cross section

0

make be = 0 or bi = 0

0'

Elliptical cross section

G

rN =

h

h2 4çæè 2 rG − 4rG2 − h 2 ÷öø Circular cross section

rG

h

make h = diameter

b 0

0' b3

Combined sections

h3

rN =

åh b r å b ln r I

I

where rN =

eI

b2

I

I

eI

I

I

h2

åh α r åα ln r

with α =

bI b1

iI

iI

re, ri: outer and inner radii

r4

h1

r3 r2 r 1

rN =

b1

0


being for the I opposite:

b

h

h1 b1 + h2 b2 + h3 b3 r r r b1 ln 2 + b2 ln 3 + b3 ln 4 r2 r3 r1

0'


page V2-4•3-3

SSM V2-4 • BEAM BENDING




Revision 0

SSM V2-4 • BEAM BENDING V2-4

4 VARIABLE HEIGHT BEAMS

A variable height beam is a beam for which a dimension measured parallel to the shear loads varies continuously over its length. Therefore, this definition excludes beams with an abrupt cross-section variation. This variation means that a correction has to be made to the longitudinal slide stress calculations, which concern the shear load(*) effectively found in a cross section: TW - bent straight beam: τ = , with W: static moment, I: Moment of inertia, b: width. Ib T W with Thv ≠ T. - bent variable height beam: τ = hv Ib (*) The normal load is not corrected owing to the smallness of the angles involved. The general assumptions are those of in-plane bending given in paragraph 2.1.1. The sign convention selected is as follows:

M>0

T>0

N>0

In practice, the following assumptions have to be added: - the cross sections involved with the area (S), are close to cross sections perpendicular to neutral fibres, - the stresses due to M and N converge at 0, - the normal component, perpendicular to (S), of these stresses is given by: σ = V2-4

4.1

N My + . S I

CORRECTION DUE TO "M" λ

T

τ1

M

f1

α1

σ(M)

h

τ1

T'

σ(M)

0

h: variable height 0: point of convergence of outer and inner fibres λ: distance of the cross section involved to point 0 M: bending moment acting on section (S) T: shear load acting on section (S) σ(M): stress due to M T': corrective shear load

α2

f1 (S)

Summing on (S) of the tangential component of the stresses due to M, gives: M T' = (tg α1 + tg α2), h value to be added algebraically to the shear load T acting on (S) M h TREDUCED = T + T' = T tg α considering α1 and α2 to be sufficiently low so that tg α1 + tg α2 ≈ tg α = ; h λ V2-4

4.2

CORRECTION DUE TO "N"

Generally, this correction is not necessary as with a bent beam the contribution of M is by far more preponderant than that of N. Revision 0 © AEROSPATIALE - 1998

Variable height beams

page V2-4•4-1

SSM V2-4 • BEAM BENDING This correction may be used when the centre of the cross section is highly offset in height in relation to point 0, a case in which the studied section is no longer normal to the neutral axis. λ h: variable height 0: point of convergence of external and internal fibres T λ: distance to point 0 of the cross section involved N N: nominal load acting on the section (S) σ(N):stress due to N σ(N) O T": corrective shear load T'' y h o τ2 βo f2 G ≠π/2 mean line (S)

Summing on (S) of the tangential component of the stresses due to N gives: T'' = N tgβo,

value to be added algebraically to the shear load T acting on (S). TREDUCED = T + T'' = T + N tgβo.

V2-4

4.3

PRACTICAL FORMULAS AND COMMENTS

1 - Equivalent height: the equation ofT' (correction due to M) gives parameter h which is the height of the cross section involved. h cannot be the geometrical height: we have seen that T' is the sum of the tangential component of normal stresses at (S). However, for a shear bowed beam, the law of distribution of normal stresses on ! (S) σ is no longer that of the longitudinal slip stresses τ. Therefore, from the shear standpoint, the beam height is considered as an equivalent height defined by: I ì I: moment of inertia of the section, Hequivalent = with í WMAX îWmax : maximum static moment.

The shear load corrected of the M effect alone is then expressed as: TREDUCED = T -

M Hequivalent

tg α, with tg α =

Hequivalent

λ

(if tg α1 + tg α2 ≈ tg α). G1

(*)

For thin web beams with a high web height , compared to that of the flanges, the height Hequivalent to be considered is the distance d separating the C.G.'sG1 and G2 of the flanges. (*) In this study, this relative magnitude criterion is true if the following checks out: I(beam total cross section) ≈ S(flanges)d2. The general equation for the shear load corrected for the effect ofM and N is: M TREDUCED = T (tg α1 + tg α2) + N tg βo. Hequivalent

d

G2

2 - Relationship between TREDUCED and T: typically, h increases with M, T' is then in the opposite direction to T and often less than it. The corrected shear load is then generally in the same direction as T and less than it, which gives the name TREDUCED. 3 - Specific case: if the resultant of the vertical loads acting on the beam is applied at 0 then T' = -T.


Variable height beams

Revision 0


CALCULATING STRESSES IN STRAIGHT BEAMS

V2-4 5.1 LINEAR ELASTICITY A straight, homogeneous and isotropic beam submitted to bending in the elastic range generates linear strains and stresses. The stress at a given point along the axis is proportional to the strain at this point and the distance of this point from the neutral fibre.

V2-4 5.1.1

Plane bending

“Elastic” plane bending occurs when the bending moment applied to a section of the beam acts around a single main axis of the section generating a maximum stress not exceeding the elastic limit. The neutral axis is then parallel to the moment axis (see fig. 2-4.5.1.1-1). +y The bending stress at a given point on the section is:

+x

Mz

Neutral axis

σf =

Mz x y Iz

σf: bending stress (> 0 if tensile).

+z

y: distance from neutral fibre along y-axis. Mz: bending moment. Iz: inertia of section around its neutral fibre.

Fig. 2-4.5.1.1-1

V2-4 5.1.2

Distribution of stresses in a beam submitted to “elastic” plane bending.

Deviated bending

“Elastic” deviated bending occurs when the bending moment, applied to a section of the beam, acts around an axis other than the main axis of inertia. In this case, the neutral fibre is no longer parallel to the moment axis. Deviated bending can be taken into account in complex section cases where the main axes are not known. However, we know that for beams submitted to bending only, the neutral axis passes through the centre of gravity of the section (see fig. 2-4.5.1.1-2). +y Neutral axis M

The neutral axis and the bending moment are in the yz plane.

+z

Fig. 2-4.5.1.1-2

Issue 0 © AEROSPATIALE

Distribution of stresses in a beam submitted to “elastic” deviated bending.

Calculating stresses in straight beams

page V2-4•5-1

SSM V2-4 • BENDING OF BEAMS The stresses can be calculated by projecting the moment onto any perpendicular axis system passing through the centre of gravity of the section under study. y

My = M.sinα

M My

α

Mz = M.cosα My, Mz: components of moment applied along y and z respectively.

z Mz

α:

Fig. 2-4.5.1.1-3

angle between the moment axis and the z-axis.

Determining stresses in a section submitted to elastic deviated bending.

The bending stress at a given point (y,z) of the section is: σf =

( M y . I yz + M z . I y ) .y I y . Iz − I

2 yz

−

( M z . I yz + M y . I z ) .z I y . I z − I 2yz

Where: I y, Iz: moments of inertia around y and z respectively. Iyz: inertia product. y, z: coordinates of the point along the y- and z-axis. Specific case: If the y- and z-axis are the main axes (this case occurs if one of the axes corresponds to a symmetry axis of the section), the inertia product Iyz is then null.

σf =

Mz . y My . z − Iz Iy

When a beam is submitted to several loading cases, it is good practice, before starting to calculate the bending stresses, to locate the main axes, the moments of inertia and the components of the various loads on these axes. Below, we recall the analytical method and the Mohr’s circle method to determine the main axes and the corresponding moments of inertia. •

Moments and inertia products along inclined axes

y

Ix1 = Ixx . cos² φ - Ixy . sin2φ + Iyy . sin² φ

y1

Iy1 = Ixx . cos² φ - Ixy . sin2φ + Iyy . sin² φ Ix1y1 = Ixy . cos² φ +

x1 ρ

+ φ

x

I xx − I yy . sin2φ 2

x1 and y1 are inclined axes. φ is the angle between x and x1.

page V2-4•5-2 © AEROSPATIALE


Issue 0

SSM V2-4 • BENDING OF BEAMS •

Main axes V

y

These are the axes where Ixx and Iyy are minimum and maximum and Ixy is null. When I xx, Iyy and Ixy are known, f, IMAX and IMIN are calculated as follows:

U ρ

+ φ

x

æ 2 I xy φ = 1 tan-1 çç 2 è I yy − I xx

IMAX =

I xx + I yy + 2

I xx + I yy + IMIN = 2 •

ö ÷ ÷ ø

é I xx − I yy ù I 2xy + ê ú 2 ë û

I

2 xy

é I xx − I yy ù +ê ú 2 ë û

2

2

Mohr’s circle Ixy

Graphic construction from values Ixx, Ixy and Iyy. (Iyy, +Ixy)

1 - determine the two coordinate points (Iyy, +Ixy) and (Ixx, -Ixy).

2φ IMIN

IMAX

Ixx, Iyy 2 - plot the circle passing through the two points the centre of which is on the Ixx, Iyy axis. 3 - IMIN and IMAX are the intersection points of the circle with the Ixx, Iyy axis. 2φ is the angle shown on the figure.

(Ixx, -Ixy)

-Ixy

I MAX + I MIN 2

Example: Calculating bending stresses at points A, B, C. 25 A B

y

Thickness of section: 2,5 mm. C

S = 237 mm²

Y

Izz = 89557 mm4

M = 105 mm.N

Iyy = 22389 mm4 α = 150° φ

Z z

Izy = - 33398 mm4 50

Caution: Mac Inertie gives Izy > 0.

25 Issue 0 © AEROSPATIALE


page V2-4•5-3

SSM V2-4 • BENDING OF BEAMS φ = 1 tan −1 2

æ 2 I xy ç ç I yy − I xx è

ö ÷ = 22.4° ÷ ø 2

I xx + I yy + IMAX = 2

I

I xx + I yy IMIN = + 2

2 xy

2 xy

é I xx − I yy ù 4 +ê ú = 103337 mm 2 ë û 2

I

é I xx − I yy ù 4 +ê ú = 8609 mm 2 ë û

Let us project moment M onto the axes (z,y)

Let us project moment M onto the axes (Z,Y)

Mz = M . cos α = 105 x cos 150° = - 86600 mm.daN My = M . sin α = 105 x sin 150° = 50000 mm.daN

Mz = M . cos (α - φ) = - 61000 mm.daN My = M . sin (α - φ) = 79230 mm.daN

Coordinates in the z,y axis system.

Coordinates in the Z,Y axis system.

ZA = - 23.75 YA = 25

ZA = zA . cos φ + yA . sin φ = - 12.4 YA = yA . cos φ - zA . sin φ = 32.2

ZB = - 23.75 YB = 22.5

ZB = zB . cos φ + yB . sin φ = - 13.4 YB = yB . cos φ - zB . sin φ = 29.9

ZC = 1,25 YC = 25

ZC = zC . cos φ + yC . sin φ = 10.7 YC = yC . cos φ - zC . sin φ = 22.6

Calculating bending stresses at points A, B and C ( M y . I yz + M z . I yy ) . y ( M z . I yz + M y . I zz ) . z σf = − I yy . I zz − I 2yz I yy . I zz − I 2yz

Calculating bending stresses at points A, B and C M . Y My . Z σf = z − Iz Iy

σf = - 4.056 . y - 8.284 . z

σf = - 0.59 . Y - 9.203 . Z

σfA = - 4.056 x 25 - 8.284 x (- 23.75) = 95.3 MPa

σfA = - 0.59 x 32.2 - 9.203 x (- 12.4) = 95.1 MPa

σfB = - 4.056 x 22.5 - 8.284 x (- 23.75) = 105.5 MPa σfC = - 4.056 x 25 - 8.284 x 1.25 = - 111.8 Mpa

V2-4 5.1.3

σfB = - 0.59 x 29.9 - 9.203 x (- 13.4) = 105.7 MPa σfC = - 0.59 x 22.6 - 9.203 x 10.7 = - 111.8 MPa

Composite beams

A beam is said to be composite if it consists of several materials with different moduli of elasticity. The strain in the elastic range is assumed to be linear and proportional to the distance between the neutral fibre and the calculation point. The elongation-stress relation is: e=σ E E = modulus of elasticity of a beam comprising several materials; it is not a constant.



Issue 0

SSM V2-4 • BENDING OF BEAMS y AiEi

y i'

G +

yi y z

n

Area:

AE =

åA E i

i =1

i

n

Centre of gravity:

åy A E i

y=

i

i

i =1 n

åA E i

i =1

i

c

Static moment:

W iE =

åy A E i =1

n

Moment of inertia:

IE =

å i =1

' i i

i

y i2 A i E i +

n

å i =1

Eoi Ei − y

2

n

åA E i =1

i

i

where: n = number of elements. c = number of the element where CG is at yi. Ioi = inertia of element i itself. As:

yi = y + y i'

Then:

IE =

n

å i =1

y i'2 A i E i +

n

åI i =1

oi

Ei

The stresses at any point i = j of the section are: Bending:

σfj=

M . E j. y 'j EI

M: bending moment

Normal:

σnj=

N . Ej AE

N: normal force

Shear:

τj=

T . W j. E j b j. IE

T: shear force bj: width of section at point j



page V2-4•5-5

SSM V2-4 • BENDING OF BEAMS V2-4 5.2 PLASTICITY V2-4 5.2.1

Plane bending

V1-5 5.2.1.1 Introduction Plane “plastic” bending is nonlinear behaviour specific to ductile materials. Indeed, these materials frequently exceed the maximum bending force calculated in linear elasticity. The characteristics of beams submitted to “plastic” bending show that the distribution of the strains remains linear whereas the distribution of the stresses is nonlinear. The latter depends on the shape of the beam and the stress-strain characteristics (figure V2-4 5.2.1.1-1). +y +x

Mz

Neutral fibre

Fig. 2-4.5.2.1.1-1

+z

Distribution of stresses in a beam submitted to plastic bending.

V1-5 5.2.1.2 Calculating bending module On the figure below, σMAX is the stress applied to the extreme fibre of a symmetrical section. The Cozzone method is based on an approximation of the profile of the real stresses; the real profile is replaced by a trapezoidal profile as shown on figure 2-4.5.2.1.2-1. σMAX

σ0

Real stress distribution

σ0 σMAX Fig. 2-4.5.2.1.2-1

Cozzone’s equivalent stress profile

Stress σ0 is a fictive stress supposed to exist at the neutral fibre of the axis. Cozzone’s method enables a fictive allowable bending stress called “bending modulus” to be calculated. This stress can be compared with the maximum stress calculated linearly. The general bending modulus expression is:


σb = σMAX + σ0 . (k - 1)


Issue 0

SSM V2-4 • BENDING OF BEAMS Remarks: Q The bending modulus calculation is based on the real required maximum stress level, that is, generally σMAX = σR at ultimate strength for a calculation at ultimate load. At limit load, the stress level must remain below the elastic limit. R The bending modulus also depends on the geometrical characteristics of the section by means of term “k”. “k” is determined by the following equation:

K=

D.W I

D

D

Where: W = static moment of the 1/2 section. I = moment of inertia of the complete section.

The values of k are given on the figure below for several symmetrical sections: K

1.0

1.5

2.0

1.33

SECTION

1.85 1.75

d

di

1.65 k

1.55 1.45

K = 16 . 3.π

1.35 1.25 0.0

0.2


0.4

0.6 d/d

0.8

1.0

æd 1 − çç i è d

ö ÷÷ ø

3

æd 1 − çç i è d

ö ÷÷ ø

4

1.2


page V2-4•5-7


e2 b

b

Section Shape Factor, K

1.70 e2/h = .5 e2/h = .4 e2/h = .3

1.60 1.50

e1

h

h

e1 e2

1.40 1.30 e2/h = .2 e2/h = .1

1.20 1.10 0.0

.2

.4

.6

.8

1.0

1.2

e1/b

K= W I/D 2 be 2 ( h − e 2 ) e 1 æ h W= + ç − e 2 ö÷ 2 2 è2 ø 2 2 be 2 e 2 + 3( h − e 2 )² e 1 ( h − 2e 2 ) 3 I/D = 12 h

(

)

b

Section Shape Factor, K

2.7

h

2.5

e1

e2

2.3 e1/b = .1 e1/b = .2 e1/b = .3 e1/b = .4 e1/b = .5 e1/b = .7 e1/b = .9

2.1 1.9 1.7 1.5 0.0

.2

.4

.6

.8

1.0

1.2

K= W I/D e h ² ( b − e 1 ) e 22 W= 1 + 8 8 e h ² ( b − e 1 )e 32 I/D = 1 + 12 12 h

e2/h



Issue 0

SSM V2-4 • BENDING OF BEAMS Let us now calculate σ0 knowing that this stress does not theoretically depend on the shape of the section. That which is valid for a rectangular section is also valid for all other sections. We wish to find σ0 corresponding to σR; this stress level is obtained at the extreme fibre of the axis. By considering pure bending and assuming that the material curve is perfectly symmetrical in tension-compression, we suppose that the stress is null at the neutral fibre of the axis. Therefore, we can plot a second x-axis corresponding to the dimension between the neutral fibre and a given fibre on the material curve (see figure 2-4.5.2.1.2-2). σ σR σ

z

z

ZMAX = h/2

σ(z)

0

z

0

Fig. 2-4.5.2.1.2-2

ε

εs z

ZMAX = h/2

Real stress profile

Thus, the bending moment can be written: h/2

ò

M = 2b z . σ( z ) . dz 0

z=h x ε 2 ε MAX Where:

dz =

Plane bending: the cross sections remain plane. ε linear with z.

h dε 2 . ε MAX

Also, ε can be written: æ ö ε = σ + 0.002 . ç σ ÷ σ E 0 . 2 è ø

n

æσ ö where: n = 500 . ç 0.2 ÷ è σR ø

n

We deduce the expressions of ε and εMAX: n

æ ö ε = σ + 1 . ç σ ÷ = εe + εp E n è σR ø n

σ æσ ö εMAX = MAX + 1 ç MAX ÷ = εeMAX + εpMAX E n è σR ø But:

σb =

M.v = 6 .M I b . h²

Moreover: σb = (k - 1) . σ0 + σMAX K=


h.W =3 I 2

Þ

σ0 æ ö = 2 . ç σb − 1÷ σ MAX è σ MAX ø


page V2-4•5-9

SSM V2-4 • BENDING OF BEAMS After resolution, we obtain: σ0 2( n − 1) = 2n + 1 σ MAX

2

æ ε P ö æ æ 2n + 1 ö æ ε e ÷÷ . ç ç . çç ÷ . çç ç è ε MAX ø è è n + 2 ø è ε P MAX

MAX

MAX

ö ö ÷÷ + 1÷ ÷ ø ø

This expression can also be written: σ0 æ n − 1 ö æç ε PMAX ö÷ ö÷ æ n − 1 ö æç ε PMAX ö÷ æç . 1−ç = 2. ç ÷. ÷.ç ÷ ç σ MAX è 2n + 1 ø çè ε MAX ÷ø ÷ø è n + 2 ø è ε MAX ø è Where: σMAX

maximum allowable stress

æσ ö εPMAX = 1 . ç MAX ÷ n è σR ø

εMAX =

n

maximum allowable plastic yielding

σ MAX + εPMAX E

æσ ö n = 500 . ç 0.2 ÷ è σR ø

maximum allowable distorsion

n

(numerical resolution)

Specific cases:

σ (σMAX = σR) Þ εPMAX = 1 ;εMAX = R + 1 n E n

(σMAX = σ0.2) Þ εPMAX = 0.2%;εMAX =

σ 0. 2 + 0.2% E

V1-5 5.2.1.3 Symmetrical sections Example: Calculation of allowable moment Y

S = 825 mm² Ix = 384444 mm4 e1 = 6 X h = 54.4

M

e2 = 6

Material: 7075 T76510 σR = 525 MPa σ0.2 = 460 MPa σ0.2c = 460 MPa E = 71700 MPa Ec = 73800 MPa n = 23 nc = 27

50



Issue 0

SSM V2-4 • BENDING OF BEAMS Step : Determine shape factor k ì e1 6 = 0.12;e 2 = 6 = 0.110 ü Þ K = 1.23 í = ý b 50 h 54.4 î þ

Step : Calculate maximum elongation The beam section submitted to a bending moment, M, has a neutral fibre at 27,2 mm from the two extreme fibres (symmetry). The maximum elongation on tensioned side is: εMAX =

σR + 1 = 525 + 1 = 0.051 E n 71700 23

We verify that εMAX < e%

0.051 < 0.07

If εMAX > e%, the limit is e%. Plane bending hypothesis: the cross sections remain plane, ε is linear with z. The elongation at the extreme fibre of the tensioned portion is 0.051. Therefore, on the extreme fibre of the compressed portion it will be: εMIN = -εMAX = - 0.051 Step : Calculate the maximum allowable stresses σMIN on compressed side and σMAX on tensioned side Let us calculate σMIN knowing that εMIN = - 0.051. σ æσ εMIN = MIN + 0.002çç MIN Ec è σ 0.2 c

ö ÷÷ ø

nc

27

- 0.051 =

σ MIN æσ ö + 0.002ç MIN ÷ Þ σMIN = - 515 MPa 73800 460 è ø

Let us check the allowable stresses of the local buckling phenomena: Local buckling of free flange (or fixed flange):

η=

1 − υe2 E s . ; 1 − υ² E c

υ=

σflt = 0.43

η x 73800 x π ² æ 6 ö 2 . ç ÷ ≈ 1687 x η MPa 12 . (1 − 0.33²) è 25 ø

Es E ö æ . υ e + çç 1 − s ÷÷ . υ P ; Es = Ec E c ø è

1 æ σ ö − 1 + 0.002 . ç flt ÷ Ec σ C 0.2 è σ C 0.2 ø

nc

σflt = - 500 MPa Local buckling of web:

σfla = 4 x

η x 73800 x π ² æ 6 ö 2 .ç ÷ ≈ 4187 x η MPa 12 . (1 − 0.33²) è 48.4 ø

After plasticity correction: σfla = - 515 MPa



page V2-4•5-11

SSM V2-4 • BENDING OF BEAMS We can therefore deduce the allowable stress in compression: σMIN = - 500 MPa The Cozzone method considers, for a symmetrical section, that σMAX = -σMIN, therefore: σMAX = 500 MPa Step : Calculate the bending modulus σb = σMAX + σ0 . (k - 1) The expression giving σ0 is: σ0 æ n − 1 ö æç ε P ö÷ æç æ n − 1 ö æç ε P ö÷ ö÷ . 1−ç = 2.ç ÷.ç ÷. ÷ σ MAX è n + 2 ø è ε MAX ø çè è 2 n + 1 ø çè ε MAX ÷ø ÷ø MAX

MAX

n

We take: ( σMAX = 500 MPa) σ0 = 405 MPa

Þ

σ æσ ö εPMAX = 1 ç MAX ÷ ;εMAX = MAX + εPMAX n è σR ø E σb = 597 MPa

Þ

Step : Calculate the allowable moment of the section Madm = σb x

Ix = 597 x 384444 = 8438 x 103 mm.N 27.2 ZMAX

V1-5 5.2.1.4 Asymmetrical sections Example: Calculating allowable moment. Y

S = 1000 mm² IXG = Ixx = 557500 mm4 IyG = Iyy = 95900 mm4

50 e2 = 8 e1 = 8 X'

X

M h = 50 38.8

Material: 7075 T76510 σR = 525 MPa σ0.2 = 460 MPa σ0.2c = 460 MPa E = 71700 MPa Ec = 73800 MPa n = 23 nc = 27

e2 = 8

25



Issue 0

SSM V2-4 • BENDING OF BEAMS Step : Divide the section above at the main axis X’X into 2 symmetrical sections Section b

Section a Y

Y

e1 = 8 X

X h = 54.4

h = 77.6

e1 = 8

38.8

e2 = 8 e2 = 8 50 25 S = 892 mm² Ix = 642400 mm4

S = 1107 mm² Ix = 472600 mm4

Step : Determine the shape factors k e ìe ü Section a: í 1 = 8 = 0.32; 2 = 8 = 0.103ý Þ K = 1.3 h 77 . 6 b 25 î þ e ìe ü Section b: í 1 = 8 = 0.16; 2 = 8 = 0.147 ý Þ K = 1.24 h 54 , 4 b 50 î þ Step : Calculate the maximum elongations. The beam section submitted to a bending moment, M, has a neutral fibre at 38,8 mm from the tensioned extreme fibre and at 27,2 mm from the compressed extreme fibre. The highest stress is obtained on the tensioned side, that is on the lower side of the beam. The maximum elongation on tensioned side is: εMAX =

σR + 1 = 525 + 1 = 0.051 E n 71700 23

We verify that εMAX < e%

0.051 < 0.07

If εMAX > e%, the limit is e%. Plane bending hypothesis: the cross sections remain plane, ε is linear with z. The elongation at the extreme fibre of the tensioned portion is 0.051. Therefore, on the extreme fibre of the compressed portion it will be: εMIN = -εMAX x 27.2 = - 0.036 38.8



page V2-4•5-13

SSM V2-4 • BENDING OF BEAMS Step : Calculate the maximum allowable stress σMIN on compressed side Let us calculate σMIN knowing that εMIN = - 0.036. σ æσ ö εMIN = MIN + 0.002çç MIN ÷÷ σ Ec è 0.2 c ø

nc

27

- 0.036 =

σ MIN æσ ö + 0.002ç MIN ÷ Þ σMIN = - 508 MPa 73800 460 è ø

Let us check the allowable stresses of the local buckling phenomena: Local buckling of free flange:

η=

1 − υe2 E s . ; 1 − υ² E c

υ=

σflt = 0.43

η x 73800 x π ² æ 8 ö 2 . ç ÷ ≈ 3000 x η MPa 12 . (1 − 0.33²) è 25 ø

Es E ö æ . υe + ç 1 − s ÷ . υ P ; Ec E c ø è

Es =

1 æ σ ö − 1 + 0.002 . ç flt ÷ Ec σ C 0.2 è σ C 0.2 ø

nc

σflt = - 514 MPa

σfla = 4 x

Local buckling of web:

η x 73800 x π² æ 8 ö 2 . ç ÷ ≈ 5184 x η MPa 12 . (1 − 0.33²) è 58 ø

After plasticity correction: σflt = - 520 MPa

We can therefore deduce the allowable stress in compression: σMIN = - 508 MPa The allowable stress on the tensioned side must be limited to a value σMAX = σR. Step : Calculate the bending moduli σb = σMAX + σ0 . (k - 1) The expression giving σ0 is: σ0 æ n − 1 ö æç ε P ö÷ æç æ n − 1 ö æç ε P ö÷ ö÷ . 1−ç = 2.ç ÷.ç ÷. ÷ σ MAX è n + 2 ø è ε MAX ø çè è 2 n + 1 ø çè ε MAX ÷ø ÷ø MAX

MAX

n

Section a:

We take: ( σMAX = 525 MPa) Þ σ0 = 474 MPa


Þ

n

Section b: We take:( σMAX = 508 MPa)

σ0 = 433 MPa


Þ

Þ



Issue 0

SSM V2-4 • BENDING OF BEAMS Step : Calculate the allowable moments for sections a and b

Section a:

Ma = σb x

Ix = 667 x 642400 = 11043 x 103 mm.N 38.8 ZMAX

Section b:

Mb = σb x

Ix = 612 x 472600 = 10634 x 103 mm.N 27.2 ZMAX

Step : Calculate the allowable moment of the real section Madm =

Ma + M b = 10839 x 103 mm.N 2



page V2-4•5-15

SSM V2-4 • BENDING OF BEAMS V2-4 5.2.2

Deviated bending

There are no accurate interaction equations concerning plastic deviated bending. The equation given below is very conservative. It is given only as a rough tool to enable the user to avoid more complex analysis methods especially when the load is not critical. When the load is critical, we suggest use of an analysis method based on the finite difference technique. We will not develop this method here. When an applied moment, M, acts around an axis other than one of the main axes, the allowable bending moment can be determined as follows: v

u θ

M

Components of the moment applied along each main axis: Mu = M . cosθ

My = M . sinθ

and

Let us determine the allowable moments Madmu and Madmv around each main axis as if the section was submitted to single bending around these axes. We can then deduce the ratios of the moments around each axis: Ru =

Mu M admu

Rv =

and

Mv M admv

Finally, let us calculate the safety factor: S.F. =

1 -1 Ru + Rv

This equation can be written in different ways especially by expressing the safety factor as a function of the components of the moment: S.F. =

1 æ cos θ M . çç + sin θ M M admv admu è

ö ÷÷ ø

-1

By writing, S.F. = 0, we can obtain the equation giving us the allowable bending moment: Madm =

1 æ cos θ çç + sin θ M admv è M admu

ö ÷÷ ø

Specific case: if components of M along u and v are equal, then: Madm =


M admu M admv = cos θ + sin θ cos θ + sin θ


Issue 0

SSM V2-4 • BENDING OF BEAMS Example: y

M Material: 7075 T76510 σR = 525 MPa σ0.2 = 460 MPa σ0.2c = 460 MPa E = 71700 MPa Ec = 73800 MPa n = 23 nc = 27

θ = 45° z

b = 20

b = 20

Step : Determine the shape factor k For a square section k = 1.5. Step : Calculate the bending modulus σb = σMAX + σ0 . (k - 1) The expression giving σ0 is: σ0 æ n − 1 ö æç ε P ö÷ æç æ n − 1 ö æç ε P ö÷ ö÷ . 1−ç = 2.ç ÷.ç ÷. ÷ σ MAX è n + 2 ø è ε MAX ø çè è 2 n + 1 ø çè ε MAX ÷ø ÷ø MAX

MAX

We take: ( σMAX = σR = 525 MPa) σ0 = 474 MPa

Þ

σ εPMAX = 1 ;εMAX = R + εPMAX n E σb = 762 MPa

Þ

Step : Calculate the allowable moments acting around the main axes as if the section was submitted to single bending around these axes Madmy = σb x

Iy y max

Square section: yMax = zMAX

and

Madmz = σb x

and

Iy = Iz

Iz z max

Where: y MAX = zMAX = 10 mm 4 Iy = Iz = 20 = 13333 mm4 12

Madmy = Madmz = 1016.103 mm.N Step : Calculate the allowable moments really acting around the main axes S.F. =

1 -1=0 Ry + Rz



page V2-4•5-17

SSM V2-4 • BENDING OF BEAMS Þ

Ry + Rz = 1

But, Ry = Rz Þ

Þ

Ry = Rz = 0.5

My = Mz = Madmy x Ry = Madmz x Rz = 1016 x 103 x 0.5 = 508 x 103 mm.N

Step : Calculate the allowable moment for the section Madm = M 2y + M 2z = 718 x 103 mm.N



Issue 0


CALCULATING STRESSES IN CURVED BEAMS

V2-4 6.1 SYMMETRICAL SECTIONS V2-4 6.1.1

Introduction

Up until now, we have considered only beams with a straight longitudinal axis before deformation. Beams where the longitudinal axes are not straight are called curved beams. The formulas established previously do not strictly apply to these beams. Below, we shall limit ourselves to giving the plane bending analysis methods for symmetrical curved beams in the elastic range. The method described in paragraph V2-4 6.1.2 disregards all distortion phenomena in the flanges. These will be studied in paragraph V2-4 6.1.3. As the method presented in paragraph V2-4 6.1.2 is simpler, the user must determine the level of accuracy required for his application then choose one of the two methods.

V2-4 6.1.2

Plane elastic bending of solid sections

When a curved beam is submitted to elastic plane bending, the distributions of the stresses and the circumferential strains are no longer linear as in straight beams but hyperbolic. These stresses increase faster on the side nearer to the centre of curvature. See figure V2-4 6.1.2.1 (where we have used arbitrary signs). The stress is proportional to the strain but, due to the difference in the lengths of the internal and external fibres of the beam, the stress and elongation are not proportional to the distance to the neutral fibre. For sections submitted to bending only, the neutral fibre does not correspond with the main axis of inertia.

Y

Main axis Neutral fibre

Z y0

M

Figure V2-4 6.1.2.1 As the stress-strain distribution of a curved beam submitted to bending is not linear, formula σx =

M.y is no longer applicable. I

The circumferential bending stress obtained at any point of the section is of following form: σx =

æ y ö M ÷ .ç S . ( rG − rN ) çè rN + y ÷ø


Calculating stresses in curved beams

page V2-4•6-1

SSM V2-4 • BENDING OF BEAMS where: M = Moment applied (positive when it decreases the radius of curvature, negative in the opposite case). S = Area of the section. rG = Radius of curvature (distance from the centre of curvature to the CG of the section). rN = Radius of the neutral fibres. y = Distance of a given point from the neutral fibre (positive in direction opposite centre of curvature). The analytical expressions of variable rN are given on page V2-4 3-3. For I- and T-sections and other shapes, refer to the method described in paragraph V2-4 6.1.3 before calculating the circumferential bending stresses.

V2-4 6.1.3

Plane elastic bending of thin profile sections

The general analysis methods described in paragraph V2-4 6.1.2 may give erroneous results, especially for I- and T-sections and other standard thin web sections due to the distortions which appear during bending. When a symmetrical curved beam is submitted to bending, the free edges of the flanges move radially and cause a reduction in the circumferential stress. This redistribution of the circumferential stresses is shown on figure V24 6.1.3.1 (a). Note that this reduction in stresses reduces the stiffness of curved beams when compared with straight beams. bs

y1

M

ea

N eu tral (a)

Figure V2-4 6.1.3.1

σr

axis

Radial stresses in web (b)

Distribution of stresses in curved beams submitted to bending

To compensate for the nonuniformity of the stresses in the flanges, we determine a flange load-carrying width called b: b = ξ . bs Where: bs = Real flange width. ξ = Load-carrying width flange factor obtained from figure V2-4 6.1.3.2. The method described in paragraph V2-4 6.1.2 can then be followed using b.



Issue 0

SSM V2-4 • BENDING OF BEAMS The deflection of the flanges generates transverse bending stresses (see figure V2-4 6.1.3.1 (a)). The transverse stress is maximum at the junction between the flange and the web. This is calledσ1: σ1 =β . σh Where: β = Transverse bending stress factor obtained from figure V2-4 6.1.3.3. σh = Circumferential bending stress at CG of the flange of the load-carrying section.

1.8

β

1.6

1.4 ba hs 1.2

1.0

rs

bs = Distance from centre of curvature to flange CG

0.8

0.6

ξ

0.4

0.0

0.5

1.0

1.5

2.0

2.5

3.0

( b s / 2)² rs . h s

Figure V2-4 6.1.3.2


Transverse bending stress β and flange efficiency factors ξ


page V2-4•6-3

SSM V2-4 • BENDING OF BEAMS In addition to the transverse stresses studied previously, radial stresses appear in the web (see figure V2-4 6.1.3.1 (b)). These are introduced by the curvature of the beam. The maximum radial stress is located at the junction between the web and the flanges. We call itσr: σr =

σ h . Ss rs . b s

Where: b a = Web thickness. Ss = Load-carrying flange area. Ri = Distances from centre of curvature to CG of upper and lower flanges, called ri and rs respectively. 30 Example:

2,5

A

M = - 50.104 mm.N 40

2,5

A

28,75

2,5 Centre of curvature

σR = 525 MPa Determine: • The maximum circumferential bending stress. • The transverse bending stress in the flange. • The maximum radial stress induced in the web. • The combined stress at the intersection of the web and the flange. • The safety factors. Step : Determine the load-carrying section The properties of the load-carrying section can be determined, for an I-section, as follows: Fixed flange:

(30 / 2)² ( b s / 2)² =3 = rs . h s ( 28,75 + 2,5 / 2) x 2,5

Þ figure V2-4 6.1.3.2: ξ = 0,42 Þ b = ξ . bs = 0,42 x 30 = 12,6 mm Free flange:

( b t / 2)² (30 / 2)² = 1,33 = rt . ht ( 28,75 + 40 − 2,5 / 2) x 2,5

Þ figure V2-4 6.1.3.2: ξ = 0,62 Þ b = ξ . bs = 0,62 x 30 = 18,6 mm



Issue 0

SSM V2-4 • BENDING OF BEAMS Load-carrying section:

18,6

S = 165,5 mm² I = 35920 mm4 2,5

G

h1 = 2,5 mm h2 = 35 mm h3 = 2,5 mm r1 = 28,75 mm r2 = 31,25 mm r3 = 66,25 mm r4 = 68,75 mm b1 = 12,6 mm b2 = 2,5 mm b3 = 18,6 mm

2,5

40

Neutral fibre rG rN 28,75

12,6


Step : Determine rN From page V2-4 3-3, expression for rN is: rN =

h1 . b1 + h 2 . b 2 + h 3 . b 3 r r r b1 . ln 2 + b 2 . ln 3 + b 3 . 4 r3 r2 r1

rN = 45,74 mm Step : Calculate the circumferential bending stress The expression for this stress at any point located at a distance y from the main axis is: σx =

æ y ö M .ç ÷ S . ( rG − rN ) è rN + y ø

The maximum stress is located at r = 28,75 mm σxMAX =

ð

y = r - rN = - 16,99 mm.

− 50.10 4 − 16,99 ö æ .ç ÷ = 379 MPa (tension). 165,5 . (50.45 − 45.74) è 45.74 + 16,99 ø

Step : Calculate the transverse bending stress factor of the real section Fixed flange:

(30 / 2)² ( b s / 2)² = =3 rs . h s ( 28,75 + 2,5 / 2) x 2,5

Þ figure V2-4 6.1.3.2: β = 1.76

Step : Calculate the circumferential bending stress at flange CG (r = 30 mmð y = - 15,74) σh =

− 50.10 4 æ ö − 15,74 .ç ÷ = 336 MPa (tension). 165,5 . (50.45 − 45.74) è 45.74 + 15,74 ø



page V2-4•6-5

SSM V2-4 • BENDING OF BEAMS Step : Calculate the max. transverse bending stress σ1 =β . σh = 1.76 x 336 = 591 MPa Step : Calculate the max. radial stress Here, this stress is located at the junction between the flange and the web. Data concerning the area of the load-carrying flange: Ss = 12,6 x 2,5 = 31,5 mm² The expression giving the radial stress is as follows: σr =

σ h . Ss 336 x 31,5 = 141 MPa = ( 28,75 + 2,5 / 2) x 2,5 rs . b a

Step : Calculate the bending stress on the upper face of the flange This stress is located at r = 31,25 mm

ð

y = r - rN = - 14,49 mm

σx = 297 MPa Step : Summary of stresses applied at web-flange junction σx = 141 MPa

x

y

z

σx = 141 MPa

σz = 141 MPa

Step : Determine the safety factors Web-flange junction:

S.F. =


2 . σR (σ x − σ y )² + (σ y − σ z )² + (σ z − σ x )²

-1

ð S.F. = 0.32


Issue 0

SSM V2-4 • BENDING OF BEAMS Extreme fibre (centre of flange lower face)

S.F. =


σR σx ² + σy ² + σx . σz

-1

ð S.F. = 0.38


page V2-4•6-7

SSM V2-4 • BENDING OF BEAMS V2-4 6.2 ASSYMMETRICAL THIN PROFILE SECTIONS V2-4 6.2.1

Introduction

Asymmetrical curved beams such as the one shown on figure V2-4 6.2.1.1 are submitted, in addition to the distortion of the flanges, to deflection of the web. The method developed below is based on: Engineering Science Data Unit, “Flange Efficiency Factors for Curved Beams Under Bending in the Plane of Curvature”, Chapter 71004.

σ1

σxt

σ1

σxt

σ1

σ1 σ1

σ1 σxs

σ1

σxs

V2-4 6.2.1.1

V2-4 6.2.2

σ1

C-section submitted to bending

Calculating load-carrying section

All references below are taken from ESDU 71004. Step : Evaluate web flexibility coefficients For each flange, determine the flexibility coefficient of the web, φ, by means of equation 4 or 5 on page 6. For certain standard sections, it will be easier to use the curves on figures 5 or 6 on pages 11 and 12. Step : Evaluate the flange efficiency factors For each flange, determine the flange efficiency factor, η (ξ in the manual), by means of equation 1 or 2 on page 6. For certain standard sections, it will be easier to use the curves on figures 1 to 4 on pages 9 and 10. Step : Determine the transverse stress ratios For each flange, determine the transverse stress ratios σ1/σxt and σ1/σxs by means of equation 1 or 2 on page 6. For certain standard sections, it will be easier to use the curves of figures 1 to 4 on pages 9 and 10. Step : Determine the load-carrying section By means of the flange efficiency factors, determine the load-carrying widths of the flanges and lips. Then, evaluate all the geometrical properties of this section such as: area, inertia, etc.



Issue 0

SSM V2-4 • BENDING OF BEAMS V2-4 6.2.3

Example

Determine the circumferential and transverse bending stresses of the section shown below and submitted to a bending moment of 70000 mm.N. 8,25 S = 30,25 mm² 2,75

Ix = 6650 mm4 40,5

70000 mm.N

0,5 449,75 Centre of curvature

Step : Evaluate the web flexibility coefficients Fixed flange:

ha (( 40,5 − 0,5) / 2) = = 0.044 rS 449,75 + 0.25 ha (( 40,5 − 0,5) / 2) = = 40 hS 0 ,5 ha 0 ,5 = =1 h S 0 ,5

From the curve on figure 5 on page 11 (ESDU), we obtain: φs = 0.53 x (1)3 x

Free flange:

40 = 3.352

ha (( 40,5 − 0,5) / 2) = = 0.0408 rt 449,75 + 40.25 ha (( 40,5 − 0,5) / 2) = = 40 ht 0 ,5 ha 0 ,5 = =1 ht 0 ,5

From the curve on figure 5 on page 11 (ESDU), we obtain:

φt = 0.52 x (1)3 x


40 = 3.289


page V2-4•6-9

SSM V2-4 • BENDING OF BEAMS Step : Determine the flange efficiency factors 2

2 b ö æ 0 ,5 ö æ ç bs − a ÷ 8, 25 − ÷ ç 2 ø 2 ø è Fixed flange: è = 0,2844 = rs . h s ( 449,75 + 0,5 / 2) x 0,5

A bt ( 2,75 − 0.25) . 0,5 = = 0.3125 0 ,5 ö b ö æ æ ÷ . 0 ,5 ç bs − a ÷ . h s ç 8, 25 − 2 ø 2 ø è è From the curves on figures 3 and 4 on page 10 (ESDU), we obtain: ξs = 0,540

σ1 = 0.560 σ xs

and

2

Free flange:

2 b ö æ 0, 5 ö æ ç bt − a ÷ 8, 25 − ç ÷ 2 ø 2 ø è è = 0,2612 = ( 449,75 + 0,5 / 2) x 0,5 rt . h t

A bt ( 2,75 − 0.25) . 0,5 = = 0.3125 ba ö 0 ,5 ö æ æ ç 8, 25 − ÷ . 0 ,5 ç bt − ÷.h 2 ø 2 ø t è è From the curves on figures 3 and 4 on page 10 (ESDU), we obtain: ξt = 0,560

and

σ1 = 0.550 σ xt

Step : Calculate the properties of the load-carrying section The area of the load-carrying section is calculated as follows:

æ ö b ö æ ö æ 0,5 ö æ Fixed flange: As = ξs . çç A bt + çç b s − a ÷÷ . h s ÷÷ = 0,540 x ç 2,5 x 0,5 + ç 8, 25 − ÷ x 0,5 ÷ = 2,835 mm² 2 2 ø è ø è è ø è ø Lip on fixed flange side:

Abt = 0,540 x (2,5 x 0,5) = 0,675 mm²

æ b æ Free flange: At = ξt . çç A bt + çç b t − a 2 è è Lip on free flange side:

ö ÷÷ . h t ø

ö æ ö 0,5 ö ÷ = 0,560 x ç 2,5 x 0,5 + æç 8, 25 − ÷ x 0,5 ÷ = 2,94 mm² ÷ 2 ø è è ø ø

Abt = 0,560 x (2,5 x 0,5) = 0,7 mm²

The load-carrying section can then be represented (see next page). Note that only the widths of the fixed flange, the free flange and the lips are modified.



Issue 0

SSM V2-4 • BENDING OF BEAMS 4,73

1,65 G

40,5

20,33 0,5 1,6

449,75


Step : Determine rN From page V2-4 3-3, expression for rN is: rN =

åh . b r å b . ln r i

i

ei

i

ii

rN = 469,68 mm Step : Calculate the circumferential bending stresses The expression for this stress at any point located at a distance y from the main axis is: σx =

æ y ö M .ç ÷ S . ( rG − rN ) çè rN + y ÷ø

The max. stress is located at r = 449,75 mm σxMAX =

y = r - rN = - 19,93 mm.

− 7.10 4 − 19,93 æ ö .ç ÷ = 304 MPa (tension). 25.53 . ( 470.08 − 469,68) è 469,68 − 19,93 ø

The min. stress is located at r = 490,25 mm σxMIN =

ð

ð

y = r - rN = - 20,57 mm.

− 7.10 4 æ ö 20,57 .ç ÷ = 288 MPa (compression). 25.53 . ( 470.08 − 469,68) è 469,68 + 20,57 ø

Step : Calculate the transverse bending stresses in fixed flange and free flange Fixed flange:


σ1 = 0.560 σ xs

ð

σ1 = ± 0.560 x σ xs


page V2-4•6-11

SSM V2-4 • BENDING OF BEAMS Where:

σxs =

− 19,68 ö æ . çç ÷÷ = 300 MPa 469 , 68 − 19 , 68 ø è

σ1 = ± 168 MPa

ð Free flange:

− 7.10 4 25.53 . ( 470.08 − 469,68)

σ1 = 0.550 σ xt

ð

σ1 = ± 0.550 x σ xt

Where:

σxt =

− 7.10 4 25.53 . ( 470.08 − 469,68)

ð

σ1 = ± 156 MPa


ö æ 20,32 . çç ÷÷ = - 284 MPa 469 , 68 + 20 , 32 ø è


Issue 0

MCS V2-5 • COLUMNS CONTENTS

V2-5 V2-5 V2-5 V2-5 V2-5 V2-5

COLUMNS 1 INTRODUCTION 2 ELASTIC ANALYSIS 3 EFFECTIVE LENGTH 4 PLASTIC ANALYSIS 5 EXAMPLE

V2-5 1

INTRODUCTION

issue

date

change

0 0 0 0 0 0

6/1999 6/1999 6/1999 6/1999 6/1999 6/1999

Creation Creation Creation Creation Creation Creation

1–1

V2-5 1.1 CONDITIONS FOR USE

1–2

V2-5 1.2 NOTATIONS AND CONVENTIONS

1–2

V2-5 2

ELASTIC ANALYSIS

2–1

V2-5 3

EFFECTIVE LENGTH

3–1

V2-5 4

PLASTIC ANALYSIS

4–1

V2-5 5

EXAMPLE

5–1

Issue 0  AEROSPATIALE

Contents

page V2-5•i

MCS V2-5 • COLUMNS


page V2-5•ii  AEROSPATIALE

Contents

Issue 0


V2-5 1 INTRODUCTION A beam is called a "column" when it is subjected to the action of two equal forces acting in opposite directions, directed towards the inside of the material, applied, along its axis, to the CGs of the end sections. The critical stress is the stress corresponding to the buckling load:

σ cr =

Fcr S

The column remains stable up to this critical load, a branch characterised by a bending strain mode corresponds to this critical load. The strain and type of failure of the columns depend mainly on: • the length of the beam. • the shape characteristics of the straight sections of the column. In this chapter, we will consider only straight columns with constant sections along their complete lengths and without a tendency for local buckling: no thin web effect (solid sections). Also, we shall take the Navier-Bernoulli hypothesis: flat cross sections before deformation, remain flat after deformation.


Introduction

page V2-5•1–1

MCS V2-5 • COLUMNS V2-5 1.1 CONDITIONS FOR USE The equations which follow are based on the flat bending theory. Also, they are applicable to solid section columns irrespective of their slenderness ratio. The clamping factor formula is applicable to all stability problems of a structural element comparable to a beam.

V2-5 1.2 NOTATIONS AND CONVENTIONS The compression stresses and compression loads are taken to be positive.

Normal load (+ compression)

Distributed linear compression load (+ compression)

Pcr = critical buckling load expressed in daN. P = compression load expressed in daN. q = distributed linear compression load expressed in daN/mm. σcr = critical buckling stress expressed in daN/mm2. Ec = modulus of elasticity in compression expressed in daN/mm2. I = moment of inertia expressed in mm4. S = area of the section expressed in mm2. K = clamping factor (dimensionless). L = column effective buckling length expressed in mm. A = real column length expressed in mm.

page V2-5•1–2  AEROSPATIALE

Introduction

Issue 0


V2-5 2 ELASTIC ANALYSIS Let us consider a beam with hinged ends.

P y y

x

L

P x

Let us calculate Pcr:

d2y Mz − P⋅ y = = 2 dx EI Gz EI Gz After resolution, we obtain:

Pcr =

π 2 ⋅ Ec ⋅ I L2

The critical stress can therefore be expressed by the Euler equation:

σ cr =

This stress can also be written:

But, the

π 2 ⋅ Ec ⋅ I S ⋅ L2

σ cr =

π 2 ⋅ Ec L2 I S

I S ratio represents the square of the radius of gyration of the section:

ρ=


I S

Elastic Analysis

page V2-5•2–1

MCS V2-5 • COLUMNS We can therefore deduce a new expression from the Euler formula:

σ cr =

π 2 ⋅ Ec (L ρ )2

The L ρ coefficient characterises the general buckling tendency. This is called the slenderness ratio and is represented by the letter λ:

λ=

L ρ

Therefore the Euler formula can be written:

σ cr =


π 2 ⋅ Ec λ2

Elastic Analysis

Issue 0


V2-5 3 EFFECTIVE LENGTH To establish the Euler formula, we used a specific case of a beam perfectly hinged at its two ends spaced apart by length L. In other end connection conditions, we agree to designate the real length of the studied beam as A. We will obtain, in each case, the distance L which should exist between two fictive hinge points to reproduce the same critical buckling load. We will call L the effective buckling length of the beam under study. The comparison of L with A will be expressed by the following general relation:

where K represents a dimensionless factor called the clamping factor specific to each case studied.

L = KA

The critical Euler load will therefore always be given by its primitive expression:

π 2 ⋅ Ec ⋅ I Pcr = L2

equivalent to

π 2 ⋅ Ec ⋅ I Pcr = K 2 ⋅ A2

The critical Euler stress will therefore be expressed by:

σ cr =

π 2 ⋅ Ec ⋅ I S ⋅ L2

equivalent to

σ cr =

π 2 ⋅ Ec ⋅ I S ⋅ K 2 ⋅ A2

The values of K for the various types of columns, boundary conditions and loads are given on figure V2-5 3.1.


Effective Length

page V2-5•3–1




Effective Length

Issue 0


V2-5 4 PLASTIC ANALYSIS The Euler formula reflects a state of elastic equilibrium: equilibrium between the bending action of P cr and the tendency of the beam to elastically straighten itself. We can therefore conclude that the Euler formula is, at maximum, valid only during the elastic period of the materials to which it is applied. When this elastic period is exceeded, we say that we have exceeded the proportionality limit, the original modulus Ec that we considered up until then and which, in reality, is valid only during the elastic period must be replaced in the Euler formula by the tangent modulus Et. We then obtain the following Euler - Engesser equations:

Pcr =

π 2 ⋅ Et ⋅ I L2

equivalent to

Pcr =

π 2 ⋅ Et ⋅ I K 2 ⋅ A2

σ cr =

π 2 ⋅ Et ⋅ I S ⋅ K 2 ⋅ A2

The Euler - Engesser critical stress will therefore be expressed by:

π 2 ⋅ Et ⋅ I σ cr = S ⋅ L2

equivalent to

For the critical stress calculation, the difficulty lies in determining the value of the tangent modulus corresponding exactly to a compression stress equal to the critical stress. Indeed, the tangent modulus Et has no fixed value but varies according to the position of the point on the stress-strain curve. This leads us therefore to make a correction to the plasticity the technique of which is described below: We call η, the plasticity correction factor:

η=

Et Ec

where

Et = tangent modulus. Ec = Young's compression modulus.

σ~cr

represents the critical stress calculated with Et = Ec Þ

η = 1.

This gives the following Euler formula:

π 2 ⋅ Ec ⋅ I σ~cr = S ⋅ K 2 ⋅ A2 If

σ~cr ≥ 0,5 ⋅ σ c 0, 2

then the critical stress after plasticity correction will be:

σ cr = η ⋅ σ~cr This calculation is iterative as


η = f (Et ) = g (σ cr ) . It converges when

Plastic Analysis

σ cr → σ~cr . η

page V2-5•4–1

MCS V2-5 • COLUMNS Bearing conditions

Type of load

K

Isolated load

1

Distributed load

0.729

Isolated load

0.5

Distributed load

0.364

Isolated load

1

Isolated load

0.7

Distributed load

0.433

Isolated load

2

Distributed load

1.12

P

A

Bi-hinged column

P

P

P Bi-clamped column, fixed transversely

A P

P

P Bi-clamped column, not fixed transversely

A P P

A

P

Column clamped at one end and hinged at the other

P

P

A

P

Column clamped at one end and free at the other

P

Figure V2-5 3.1: Clamping factors (continued on next page)


Plastic Analysis

Issue 0


Bearing conditions

Type of load

K

Isolated load

See figure below

P W W

Bi-hinged column on elastic supports

A W W

P 11 10,5 10 9,5 9 8,5 8 7,5 7

(π/K)2

6,5 6 5,5 5 4,5 4 3,5 3 2,5 2 1,5 1 0,5 0 0

2

4

6

8

10

12

14

16

18

20

22

24

26

28

30

32

34

36

38

40

42

44

46

48

50

52

54

56

58

60

k.W3/EI

k represents the rigidity of an elastic intermediary support. This graph is valid for a number of supports tending towards infinity but we can assume that it remains valid in all cases.


Plastic Analysis

page V2-5•4–3

MCS V2-5 • COLUMNS Bearing conditions

Type of load

K

P-qL

A

qL

Bi-hinged column, fixed transversely

Isolated load + distributed load

1 − 0,47 ⋅

Bi-clamped column, fixed transversely


0,25 − 0,12 ⋅

qL P

Column clamped at one end and hinged at the other end


0,49 − 0,3 ⋅

qL P

Column clamped at one end and free at the other end


4 − 2,74 ⋅

qL P

P

P-qL

A

qL

P

P-qL

A

qL

P

P-qL

A

qL

qL P

P

Figure V2-5 3.1: Clamping factors.


Plastic Analysis

Issue 0


V2-5 5 EXAMPLE Data: S = 200 mm2. I = 1667 mm4.

P

Material: 2024 T351 Ec = 7380 daN/mm2. σR = 44 daN/mm2. σc0.2 = 32.5 daN/mm2. nc = 8.

150

P Question: Calculate the critical buckling stress. Result: Step 1: Determine K K = 0.7 (see figure V2-5 3.1). Step 2: Calculate the critical buckling stress We use the Euler formula:

π2 ⋅ Ec ⋅ I π 2 ⋅ 7380 × 1667 ~ σ cr = = = 55 daN / mm 2 2 2 2 2 S⋅ K ⋅ A 200 × 0,7 × 150 Step 3: Apply a plasticity correction We call

σ~cr

the critical stress calculated with η = 1,

σ cr = η ⋅ σ~cr Where:

η=

Et Ec

We use the Ramberg and Osgood model: nc ì æ σ cr ö σ cr ÷ ïε = + 0,002 ⋅ çç ÷ Ec ï è σ c 0,2 ø ï σ cr ï íEs = ε ï nc 1 − nc ï1 ïE = E + E s c ï t î

We obtain, after numerical resolution:

σ cr = 27,1 daN / mm 2 Issue 0  AEROSPATIALE

Example

page V2-5•5–1

MCS V2-6 • BEAMS-COLUMNS CONTENTS issue

date

change

0 0 0 0 0

6/1999 6/1999 6/1999 6/1999 6/1999

Creation Creation Creation Creation Creation

V2-6 BEAMS-COLUMNS V2-6 1 INTRODUCTION V2-6 2 CALCULATION OF AMPLIFIED BENDING MOMENT V2-6 3 ALLOWABLE STRESS V2-6 4 EXAMPLE

V2-6 1

INTRODUCTION

1–1

V2-6 1.1 CONDITIONS FOR USE

1–2

V2-6 1.2 NOTATIONS AND CONVENTIONS

1–2

V2-6 2

CALCULATION OF AMPLIFIED BENDING MOMENT

2–1

V2-6 3

ALLOWABLE STRESS

3–1

V2-6 4

EXAMPLE

4–1


Introduction

page V2-6•i

MCS V2-6 • BEAMS-COLUMNS


page V2-6•ii  AEROSPATIALE 1999

Introduction

Issue 0

MCS V2-6 • BEAMS-COLUMNS V2-6 1

INTRODUCTION

A beam is called a "beam-column" when it is submitted both to compression and bending loads. When a structural element similar to a beam is subjected, simultaneously, to a normal load and a bending moment, we cannot use the superposition theorem to determine the stresses of which it is the seat. Indeed, a part of the bending moment is due to the action of the normal load on the deflected beam. Thus, a normal tensile load the direction of which is constant and therefore parallel to the axis of the nondeflected beam tends to flatten the deflection. Under the same conditions, a compression load tends to amplify it. In this chapter, we have not considered the case where the normal load is turning (remaining tangent to the deflected beam). Let us simply invoke on this point the case of the Beck beam (one end clamped, the other end free, compressed by an external force which remains tangent to the medium fibre). This system has no critical buckling load but a branch point characterised by a transition to an oscillating dynamic state. From a practical viewpoint, we apply the superposition theorem to the bent and tensioned beams as this approach is conservative. However, for beams-columns, we must calculate the amplified bending moments. The verification of the good resistance of a beam-column will include two steps: - The study of the column as described in § V2-5. - The study of the beam-column if the axial load is lower than the critical load calculated in the previous step.


Introduction

page V2-6•1–1

MCS V2-6 • BEAMS-COLUMNS V2-6 1.1

CONDITIONS FOR USE

See chapter V2.5 "Columns".

V2-6 1.2

NOTATIONS AND CONVENTIONS

The compression loads are taken to be positive. A positive bending moment compresses the upper fibres of the beam. A distributed lateral load leads to a positive moment. A positive lateral load compresses the upper fibres of the beam. The compression stresses are positive.

Bending moment at end (+ compresses upper fibres)

Bending moment at origin (+ compresses upper fibres)

Normal load (+ compression) Distributed lateral load (+ compresses upper fibres)

P = compression load expressed in daN. q = distributed lateral load expressed in daN/mm. F = applied isolated load expressed in daN. σcr = critical buckling stress expressed in daN/mm2. Ec = modulus of elasticity in compression expressed in daN/mm2. I = moment of inertia expressed in mm4. S = area of the section expressed in mm2. K = clamping factor (dimensionless). L = column effective buckling length expressed in mm. A = real column length expressed in mm.

page V2-6•1–2  AEROSPATIALE 1999

Introduction

Issue 0

MCS V2-6 • BEAMS-COLUMNS CALCULATING AMPLIFIED BENDING MOMENT

V2-6 2

Let us consider a beam where the ends are on single supports with compression loads (Nx) and bending loads (Mz).

y y P

Mz B

Mz A

P B

A

x

x L Beam with compression and bending loads The bending moment at x is:

æ Mz B − Mz A ö Mz = Mz A + ç ÷⋅x − P⋅y L è ø Let us differentiate / x:

d2y d 2 Mz = −P ⋅ 2 2 dx dx

where:

Therefore:

d 2 Mz P + ⋅ Mz = 0 EI dx 2

Let us write:

j=

d 2 y Mz = EI dx 2

EtI P

The solution of this equation is: Mz = C1 ⋅ sin At x=0: Mz = Mz A

x x + C 2 ⋅ cos j j

C 2 = Mz A Mz B − Mz A ⋅ cos

At x = L: Mz = Mz B :

C1 = sin

L j

L j

We therefore obtain: æ æ öö ç Mz B − Mz A ⋅ cosç L ÷ ÷ ç j ÷÷ ç è ø ⋅ sin æç x ö÷ + Mz ⋅ cosæç x ö÷ Mz = ç A ÷ ç j÷ ç j÷ æ Lö è ø è ø ÷ ç çç ÷÷ sin ÷ ç j è ø ø è


Calculating Amplified Bending Moment

page V2-6•2–1

MCS V2-6 • BEAMS-COLUMNS Mz is maximum for:

dMz =0 dx æ xö æ xö C C Þ 1 ⋅ cosçç ÷÷ − 2 ⋅ sin çç ÷÷ = 0 j j è jø è jø That is for tan

x C1 = j C2

Mz maximum for: Lö æ ç Mz B − Mz A ⋅ cos ÷ j÷ x = j ⋅ Arc tanç ÷ ç L sin ÷ ç j ø è

By studying other examples, we can see that the bending moment expression is always in following form:

Beam - column bending moment: Mz = C1 ⋅ sin

x x + C2 ⋅ cos + j 2 ⋅ f ( x ) j j

Values of C1 , C2 and f(x) are given on figure V2-6 2.1. Remark: it is possible to combine several cases (see example below).

M1

Case ①

M2

M 2 − M 1 cos P

P L

✚

L sin j

L j

⋅ sin

x x + M 1 ⋅ cos j j

✚

✚

q

Case ②

P

P

j 2 ⋅ q ⋅ tan

x L x ⋅ sin + j 2 ⋅ q ⋅ cos − q ⋅ j 2 2j j j

L

=

= M1

Case }

=

q

P

P L


M2

Lö æ M2 − M1 ⋅ cos ÷ ç x x L j÷ ç j2 ⋅ q ⋅ tan + ⋅ sin + M1 + j2 ⋅ q ⋅ cos − q ⋅ j 2 ç ÷ L 2j j j sin ç ÷ j è ø


(

)

Issue 0

MCS V2-6 • BEAMS-COLUMNS Conditions

C1

① M1

M 2 − M 1 cos

M2 P

P

L j

L j

sin

L

C2

f(x)

M1

0

j2 ⋅ q

-q

j ⋅ qL L 2 tan 2j

-q

0

0

②

q P

P

L 2j

j 2 ⋅ q ⋅ tan

L

③

q

j ⋅ qL 2

P

P L x≤a a
a

P

x

④

F

j ⋅ F sin

b P

sin

L x>a a
a

L j

⑤

F

j ⋅ F sin

b x

P

−

P

tan

L

M2

j2 ⋅ q ⋅ tan

P

P

a

P

x

L + 2j

M2 − M1 ⋅ cos

M ⋅ cos

b M

−

P

sin

P

L j

L sin j

⑦

L x>a a
j ⋅ F sin

L j

L x≥a a
a j

}

q

M1

b j

b j

L j

a j

2 M1 + j .q

0

0

-q

0

⑧ a x

M ⋅ cos

b M L

P

− tan

a j

L j

M ⋅ cos

a j

0

Figure V2-6 2.1 (continued on next page)



page V2-6•2–3

MCS V2-6 • BEAMS-COLUMNS Conditions

C1

C2

f(x)

j2.q

q ⋅ ( x − 1) L

j2 ⋅ q L sin j

0

−q ⋅ x L

2 j2 ⋅ q L L ⋅ cos j 2j

0

⑨

q

− P

P L

⑩

q P

P L x ≤ L/2

P

11

L/2

q

x

P L

x≤a a
a

P

x

P

−

2q ⋅ x L

12

F

L b b æb − ⋅ sin ç ⋅ cos j j j j j ⋅ Fç ç L L L ⋅ cos − sin ç j j è j

b P L

x>a a
j2 ⋅ q L tan j

ö ÷ ÷ ÷ ÷ ø

L b b æb − ⋅ sin ç ⋅ cos j j j j j ⋅ Fç ç L L L ⋅ cos − sin ç j j è j

ö ÷ ÷ ÷ ÷ ø

0

13

F a

b x

P L

é Læ a L a b öù ê cos çç sin − cos + ÷÷ ú j è j j j j øú j ⋅ Fê ê ú L L L cos − sin ê ú j j j êë úû

é L æ a L a b öù cos − ÷÷ ú ê sin çç − sin + j è j j j j øú j ⋅ Fê ê ú L L L − sin cos ê ú j j j ëê ûú

0

L L ö æ − ÷ ç tan L 2 j 2 j÷ ç ⋅ j ⋅ q ⋅ L + j 2 ⋅ q ⋅ tan ç L L ÷ 2j ÷ ç tan − j j ø è

L L ö æ + ÷ ç − tan L 2 j 2 j÷ ç ⋅ j ⋅ q ⋅ L ⋅ tan + j 2 ⋅ q ç L L ÷ j ÷ ç tan − j j ø è

-q

14

q

P

x

P L 15

q

j.qL P

P L

j

2

ö æ ÷ ç L 1 L ç ⋅q − tan ÷ ç L j÷ j ÷ ç cos j ø è

-q

16

F

j.F P

P

− j ⋅ F ⋅ tan

L j

0

L Figure V2-6 2.1



Issue 0


ALLOWABLE STRESS

Two main cases must be considered when choosing allowable stress:

♦

Solid beams

No local buckling phenomena occur before general buckling, the allowable stress on tensioned side is equal toσR. We will then consider that the allowable stress on compressed side is also equal to σR.

♦

Thin web beams

A local buckling phenomenon can occur before general buckling. This state does not give rise to an immediate collapse but can considerably decrease the allowable breaking stress. The allowable stress on compressed side is equal to:

σ adm = MIN{σ local buckling ;σ R }. The allowable stress on tensioned side is equal to:

σ adm = σ R . Summary:

Tensioned section: Compressed section:


σ adm = σ R . Solid beams

:

σ adm = σ R .

Thin web beams

:

σ adm = MIN σ local buckling ; σ R .

Allowable Stress

{

}

page V2-6•3–1


EXAMPLE

Data:

A = 100 mm. a = 30 mm. c = 10 mm. S = 100 mm2. I = 833 mm4. F = 70 daN. P = 1160 daN. M1 = 1200 mm.daN. M2 = 4000 mm.daN.

c

D

F M1 P

A

M2 a

B

C D

c P

A

Material: 7075 T76510 Ec = 7380 daN/mm2. nc = 27. σc0,2 = 46 daN/mm2. σR = 52.5 daN/mm2.

DD

E = 7170 daN/mm2. n = 23. σ0,2 = 46 daN/mm2. e% = 7.

Question: Calculate the maximum amplified bending moment and the Reserve Factor. Result:

Step 1: Calculate the allowable compression stress There is no local buckling phenomenon as the beam is solid in section therefore σR represents the allowable compression stress.

σ adm = σ R = 52.5 daN/mm2. Step 2: Column study (see §V2-5) We check that the beam does not buckle under the action of P alone. • Determine K K = 1 (see figure V2-5 3.1).

• Calculate the critical buckling stress We use the Euler formula: 2 2 ~ = π ⋅ E c ⋅ I = π ⋅ 7380 × 833 = 60,7daN / mm 2 σ cr 2 2 S⋅ K ⋅ A 100 ×12 ×100 2

• Apply a plasticity correction We call σ~cr the critical stress calculated with η = 1,

σ cr = η ⋅ σ~cr Where:

η=


Et Ec

Example

page V2-6•4–1

MCS V2-6 • BEAMS-COLUMNS We use the Ramberg and Osgood model: ì æ ïε = σ cr + 0,002 ⋅ ç σ cr çσ ï Ec è c 0, 2 ï σ ï cr íEs = ε ï 1 − nc ï1 n = c + ï Es Ec ï Et î

ö ÷ ÷ ø

nc

We obtain after numerical resolution:

σ cr = 41,1 daN / mm 2 But, the stress applied due to the axial load is: σ=

P = 11,6 daN / mm 2 S

As σ < σ cr , we can deduce that the column does not buckle. Step 3: Determine the maximum bending moment expressions j=

EtI = P

7380 × 833 = 72,8mm as Et = Ec. 1160

• For x ≤ a: By using a combination of cases 1 and 4 (see figure V2-6 2.1), we obtain:

M 2 − M 1 ⋅ cos C1 =

A j

j ⋅ F ⋅ sin

+ A A sin sin j j C2 = M1 = 1200 mm.daN.

b j

= 8101 mm.daN.

f(x).j2 = 0.

M AC = 8101 ⋅ sin

x x + 1200 ⋅ cos j j

• For x > a: By using a combination of cases 1 and 5 (see figure V2-6 2.1), we obtain: A A j⋅ F ⋅ sin M 2 − M 1 ⋅ cos j j C1 = − = 3431 mm.daN. A A tan sin j j C2 = M 1 + j⋅ F ⋅ sin


A = 3241 mm.daN. j

Example

Issue 0

MCS V2-6 • BEAMS-COLUMNS f(x).j2 = 0.

M CB = 3431 ⋅ sin

x x + 3241 ⋅ cos j j

Step 4: Determine the maximum bending moment • At point A:

M A = 8101 ⋅ sin 0 + 1200 ⋅ cos 0 = 1200 mm.daN. • At point C:

M C = 8101 ⋅ sin

30 30 + 1200 ⋅ cos = 4344 mm.daN. 72,8 72,8

• At point B:

M B = 3431 ⋅ sin

100 100 + 3241 ⋅ cos = 4000 mm.daN. 72,8 72,8

• Between A and C:

dM AC 8101 1200 x x = ⋅ sin − ⋅ cos =0 dx 72,8 72,8 72,8 72,8 The moment is maximum at abscissa point:

x = 72,8 ⋅ Arc tan

8101 = 103.7 mm. 1200

But: 103.7 > xC. Therefore MAC is maximum at xc. MAC max = 4344 mm.daN. • Between C and B:

dM CB 3431 3241 x x = ⋅ sin − ⋅ cos =0 dx 72,8 72,8 72,8 72,8 The moment is maximum at abscissa point:

x = 72,8 ⋅ Arc tan

3431 = 59.3 mm. 3241

M CB max = 3431 ⋅ sin

59,3 59,3 + 3241 ⋅ cos = 4720 mm.daN. 72,8 72,8

Therefore, the maximum moment on beam AB is: MAB max = MAX ( MAC max ; M CB max ) = 4720 mm.daN


Example

page V2-6•4–3

MCS V2-6 • BEAMS-COLUMNS Step 5: Calculate the total stress and the Reserve Factor • Stress due to the axial load (see step 1): σc =

P = 11,6 daN / mm 2 S

• Stress due to the bending moment:

σf =±

M c ⋅ = ± 28,3 daN / mm 2 I 2

• Maximum compression stress:

σ MAX = σ c + σ f = 39.9 daN/mm2. • Reserve Factor:

R.F . =


52,5 = 1,32 39,9

Example

Issue 0

Static Metallic Manual - Mts004

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