Structural Substantiation Report · 2017-06-16 · existing aircraft skin with NAS1097AD4 rivets. This report is providing the structural substantiation static, fatigue, and damage

Form KNSI-P3-004-7 Issue: 0 Rev: 0 Structural Substantiation Report

DOA EASA.21J.560

STRUCTURAL SUBSTANTIATION REPORT

Document number: 15K036-SSR-001-1.R Page 1 of 109

Page 1 of 109

Part 1: General Description

TITLE: FUSELAGE - VHF ANTENNA INSTALLATION

VALID ON AIRCRAFT:

TAB No. Serial Number Registration

B757-200 24868 VQ-BOX

REVISION STATUS:

Orig: February 26, 2015

Rev: 01 June 25, 2015

Description: This is Structural Substantiation Report provides addition substantiation data to validate the

modification approved by the KNSI Classification and Certification Sheet 15K036-CCS-004-0.R.

Structural Substantiation Report

This document and all information and expression contained herein are the property of KNSI Limited and

are provided to the recipient in confidence. This document contains proprietary information and shall at all times remain the property of KNSI Limited, no intellectual property right or licence is granted by KNSI

Limited in connection with any information contained herein and the information contained herein shall be

treated as confidential and not disclosed to any third party without the prior written consent of KNSI Limited

Part 2: Approval

Prepared By:

Compliance Verification

Engineer:

Office of Airworthiness:

Name: Aruni Senanayaka Name: Y. Dissanayake Name: K. Obeysekara

Date: June 25, 2015 Date: June 25, 2015 Date: June 25, 2015


DOA EASA.21J.560



Page 2 of 109

Revision status

Rev 00: Initial Issue - February 26, 2015

Rev 01: Antenna location is changed to the bottom of the aircraft – June 25, 2015

General Introduction

The VHF antenna at the tail section of the aforementioned Boeing 757-200 aircraft is going to be changed

according to the KNSI change bulletin 15K036-CB-004-0.R or the latest revision.

This Structural substantiation report was raised to substantiate this change.

This report is providing the structural substantiation static, fatigue, and damage tolerance for alterations

made to a Boeing 757-200 by the installation of the VHF Antenna and transceiver.

Reference Documents

a) Structural Analysis Report No. LB-VHF.757-703SA

b) KNSI Drawing 15K036-MD-001-0.R - VHF Antenna Installation

Description

The -01 VHF Antenna installation is detailed in KNSI Drawing 15K036-MD-001-0.R. The antenna is

installed on the bottom of the aircraft at F.S. 1452 between STR 30 and STR 29R. The antenna is

attached through the aircraft skin to the -11 hat section using ten ¼-28 fasteners into nutplates attached to the -11 hat section. The -11 hat section is attached to two aft -1 5 mount channels and two -13 FWD

mount channels using three MS20470AD5 rivets per channel. The -11 hat section is also attached to the

-17 stringer support using four HL18-5 Hi-Loks. The -17 stringer support is attached to the -19 and -20 angle supports using two MS20470AD5 rivets per angle. The -19 and -20 angle supports are attached

to the existing stringers using two MS20470AD6 rivets each. The -15 channels are attached to the

existing frame web at F.S. 1 460 using two MS20470AD5 rivets each. The -13 FWD channels are attached to the existing frame web at F.S. 1440 using two MS20470AD5 rivets each. The installation drills one

1.31-in x 1.71-in feed thru hole in the 0.070-in thick skin for the antenna installation. To restore strength

to the fuselage skin the internal 0.080 inch thick 2024-T3 aluminum -11 hat section is attached to the existing aircraft skin with NAS1097AD4 rivets.

This report is providing the structural substantiation static, fatigue, and damage tolerance for alterations

made to a Boeing 757-200 by the installation of the VHF Antenna and transceiver. Refer to KNSI drawing

15K036-MD-001-0.R for details.

Inspection Intervals

The following lists the threshold and recurrent inspection intervals for the Fatigue Life Evaluation and

each crack growth model of the Damage Tolerance Evaluation (DTE). The inspections intervals listed do not change the existing inspection or maintenance program requirements for the aircraft unless the

intervals listed below would occur prior to an existing inspection.

Skin:

For Lateral Cracks:

THRESHOLD INSPECTION (based on DSG) = 25,000 Cycles (See Page No. 46) RECURRING INSPECTIONS (based on DSG) = 12,500 Cycles (See Page No. 46)

FATIGUE LIFE (based on DSG) = 50,000 Cycles (See Page No. 46)

For Longitudinal Cracks: THRESHOLD INSPECTION (based on DSG) = 25,000 Cycles (See Page No. 70)

RECURRING INSPECTIONS (based on DSG) = 12,500 Cycles (See Page No. 70)

FATIGUE LIFE (based on DSG) = 50,000 Cycles (See Page No. 70)


DOA EASA.21J.560



Page 3 of 109

Stringers:

THRESHOLD INSPECTION = 25,000 Cycles (See Page No. 88)

RECURRING INSPECTIONS = 12,500 Cycles (See Page No. 88) FATIGUE LIFE = 50,000 Cycles (See Page No. 88)

-11 Hat Section:

THRESHOLD INSPECTION = 25,000 Cycles (See Page No. 94) RECURRING INSPECTIONS = 12,500 Cycles (See Page No. 94)

FATIGUE LIFE = 50,000 Cycles (See Page No. 94)

Frames:

The 0.159-in holes drilled into the existing frames for the installation of new MS20470AD5 rivets are the

same size and have the same edge distance as other rivet holes in the frames. Therefore the inspection

schedule for the existing adjacent rivet holes shall also apply to the new rivet holes in the frames.

The above inspection intervals are based upon a High Frequency Eddy Current (HFEC) Inspection, (See

Table 5.0.4A); refer to the specific maintenance instructions for inspection details.

Damage Tolerance Assessment (DTA) Method

The Damage Tolerance Assessment (DTA) is performed in accordance with the guidance provided by the

Seattle Aircraft Certification Office (SACO) and Patrick Safarian, dated October 1999. The following

outline lists the steps required, to determine the inspection intervals for the repair or alteration, to

support continued airworthiness.

a. Select structures that require DTA to establish special inspections

The fuselage skin and doubler are considered Fatigue Critical Structure and require a DTA.

b. Obtain fatigue loads

The fatigue loads are obtained through calculation, as required by 14 CFR Part 23.571.

c. Develop flight profile

The flight profile is considered as the flight cabin pressurization cycle

d. Develop exceedance spectrum

The exceedance spectrum consists of a single pressurization cycle (one flight equal to one cycle).

e. Develop stress spectrum for each individual structure/area to be analyzed

The stress spectrum for the skin is considered for longitudinal loading and lateral loading. The spectra

are single cycle constant amplitude.

f. Establish the initial flaw sizes, to be assumed

The initial flaw sizes are established, 0.05 inches for an initial crack at a hole, and 0.010 inches for

continuing damage.

g. Determine stress intensity factors for cracking scenarios to be addressed

The stress intensity factors and crack growth properties are addressed through the use of the AFGROW program and the NASGRO Equation are determined.

h. Obtain material properties for crack growth calculations and residual strength analyses

The material properties for the crack growth and residual strength analyses are obtained from the AFGROW material database.

i. Determine required residual strength loads

The required residual strength for the longitudinal and lateral loading conditions are determined.

j. Calculate residual strength and critical crack length


DOA EASA.21J.560



Page 4 of 109

The residual strength and associated critical crack lengths are calculated.

k. Determine detectable crack length as a function of inspection method

The detectable (inspectable) crack lengths as a function of inspection method are determined as listed in Table 5.0.4A.

l. Calculate crack growth life

The crack growth life for the rivet rows and feed through holes are calculated using the AFGROW

program and described.

m. Determine inspection thresholds

The threshold inspections are described and calculated. A summary of the threshold inspections is

presented at the beginning of this document prior to the introduction.

n. Determine repeat inspections intervals

The recurring inspections are described and calculated. A summary of the recurring inspections is

presented at the beginning of this document prior to the introduction.

o. For modifications develop Instructions for Continued Airworthiness, and for repairs update

maintenance program to incorporate new special inspections.

The Instructions for Continued Airworthiness (ICA) and maintenance program updates incorporating the new special inspections are not contained within this report.


DOA EASA.21J.560



Page 5 of 109

Weight and Load Factors

VHF Antenna =4.0 lbs

Other Hardware <2.5 lbs

Total Antenna Installation Weight = 4.0 + 2.5 = 6.5 lbs

Based upon the profile of this component, there are lift and drag loads considered in the longitudinal

and lateral directions.

The antenna is attached directly to the -11 doubler. The -11 hat section supports the out-of plane loads

of the antenna due to aerodynamic loads. However, the in-plane loads are supported by the skin/doubler

in bearing.

Cabin Pressurization

The maximum cabin pressurization setting is considered to be the maximum relief valve setting (MRVS)

pressure. This pressure in combination with a relief valve maintenance tolerance is considered the

normal operating differential pressure (ΔP). Per 757-200 operating manual, the maximum relief valve setting is 8.95 psi.

ΔP = (8.95 psi + 0.5 psi*) = 9.45 psi

*Ref valve maintenance tolerance


DOA EASA.21J.560



Page 6 of 109

ΔPLIM = (9.45 psi + 0.5 psi*)(1.33**) = 13.23 psi

*Ref external aerodynamic effect

**Ref Limit Load Factor

ΔPULT = (13.23 psi)(1.5*) = 19.85 psi

*Ref ultimate load factor

Vertical and Gust & Maneuver Load Factors

Fatigue Analysis Vertical Load Factor (not aircraft specific)

Nz (ult) = 3.80 g (Down) Nz (lim) = 3.80 g/1.5 = 2.54 g (Down)

Emergency Landing Condition

NZ (ult) = 3.0 g (Down)

NY (ult) = 1.50 g (Side)

NX (ult) = 9.00 g (Forward)

Fitting Factor

F.F. = 1.15

The following aircraft axis and sign conventions are used throughout this report.

Figure No. 3.0.2A: Aircraft axis and sign conventions

Aerodynamic Loads

Drag load

From Type Certificate Data Sheet A2NM, VMO = 350 knots KCAS

For VMO = 350 knots at sea level (at sea level indicated airspeed equals true airspeed):

VMO = (350 knots) (1.151 mile/knot) / (3600 sec/hr / 5280 ft/mile) = 590.85 ft/sec

VNE = 590.85 / 0.8 = 738.6 ft/sec


DOA EASA.21J.560



Page 7 of 109

This airspeed results in the dynamic pressure, q at sea level, which is calculated as:

= air density at sea level = 0.002377 lb.-sec2/ft4

qNE = ½ (Vd)2 = ½ (0.002377 lb.-sec2/ft4)(738.6 ft/sec)2 = 648.4 lb/ft2

qMO = ½ (VMO)2 = ½ (0.002377 lb.-sec2/ft4)(590.85 ft/sec)2 = 414.91 lb/ft2

The frontal area of the antenna is estimated using AutoCAD.

Figure No. 3.0.3A: Area of the antenna

The drag coefficient of a streamline body like the antenna: Cd = 0.12

Figure No. 3.0.3B: Coefficient of Drag Diagram

For this analysis conservatively consider a drag coefficient of 0.24

This results in a maximum drag force of:

Fx = Cd qNE A = (0.24)[(648.4 lb/ft2)/(144.0 in2/ft2)](13.725 in2) = 14.83 lbs

Side Lift load

The manoeuvrings loads that would produce a yaw angle of 10 degrees are designed to withstand the

loading produced by the control surface at VA. The load condition for the side aero load of the antenna

is considered to be at VMO and at a yaw angle of 10 degrees.


DOA EASA.21J.560



Page 8 of 109

The side area of the antenna is estimated using AutoCAD:

Figure No. 3.0.3C: The side area of the antenna

The lift coefficient of a streamline body like the antenna:

CL = 1.0 @ 10° angle of attack*

*Ref. NACA 64A010 Wing Section

This results in a maximum lift force of:

Fy = CLqMO A = (1.0)[(414.91 lb/ft2)/(144.0 in2/ft2)](153.0085 in2) = 440.87 lbs

Fatigue Loads

14 CFR Paragraph 25.571(b)(5) amendment 25-45:

(i) The normal operating differential pressure combined with the expected external

aerodynamic pressures applied simultaneously with the flight loading conditions specified in

paragraphs (b)(1) through (4) of this section, if they have a significant effect.

(ii) The expected external aerodynamic pressure in 1g flight combined with a cabin differential

pressure equal to 1.1 times the normal operating differential pressure without any other load.

General Assumptions:

* The antenna is on the skin bay at STA 1440 and STA 1460 between STR 28L and STR 29L.

* Biaxial loading due to pressure plus vertical inertia fuselage bending only, inertial shear is neglected.

* The beneficial effects of frames and stringers are ignored.

* Two conditions must be considered: Condition (i) Normal pressure combined with limit flight loads,

Condition (ii) Factored pressure loading.

* The lower skin is in compression during the flight. However, the skin may experience tension loads during landing.

Residual Strength

Condition (i) Residual Strength Stress:


DOA EASA.21J.560



Page 9 of 109

σ LongRes=(ΔP+ 0.5 psi*)(R)/[(2)(t)]+(Nz)(σ1g,max) (aft of

front spar) *(Ref external aerodynamic pressure)

σ LongRes=(ΔP+ 0.5 psi*)(R)/[(2)(t)]+ (L/S)**(Nz)(σ 1g,max) (forward of front spar)

*(Refexternal aerodynamic pressure)

**(Ref. Figure 3.1A)

Condition (ii) Residual Strength Stress:

σ HoopRes= [(1.1)(ΔP) + 0.5 psi*](R)/(t)

* (Refexternalaerodynamic pressure)

LEFM

Condition (i) Cyclic LEFM Stress:

σ LongLEFM=(ΔP)(R)/[(2)(t)]+ (1.5)(σ 1g,max) (aft of front spar)

σ LongLEFM=(ΔP)(R)/[(2)(t)]+ (L/S)* (1.5)(σ 1g,max)

(forward of front spar)

*(Ref. Figure 3.1A)

Condition (ii) Cyclic LEFM Stress:

σ HoopLEFM=(ΔP)(R)/(t)


DOA EASA.21J.560



Page 10 of 109

Condition (i) Far Field Stress (Longitudinal Direction):

To calculate the stresses for Condition (i) the σ1g,max stress must be calculated.

The original skin thickness is 0.070-inch at the alteration.

The radius is considered to be the average of the larger and smaller radii.

Average fuselage radius = (78.0 in + 84.0 in) /2 = 81.0 in

σR-axial = [1.1(ΔP)+ 0.5 psi*](R)/[(2)(ts)]

= [1.1(9.45) + 0.5]psi(81.00 in)/[(2)(0.070 in)] = 6,304 psi

*Ref external aerodynamic pressure


DOA EASA.21J.560



Page 11 of 109

The antenna location is located on the fuselage section lower skin (below the neutral axis of the airplane)

and therefore, does not experience tension stresses due to fuselage bending. Only the longitudinal

stresses due to pressure are applicable for the antenna location.

Condition (i) Residual Strength Stress

σLongRes = 6,304 psi

Condition (i) Cyclic LEFM Stress

σLongLEFM = (9.45 psi)(81 .00 in)/[(2)(0.070 in)] = 5,468 psi

Condition (ii) Far Field Stress (Lateral Direction):

Condition (ii) Residual Strength Stress

σHoopRes = [1.1(9.45 psi) + 0.50 psi*](81.00 in)/(0.070 in) = 12,607 psi

*Ref external aerodynamic pressure

Condition (ii) Cyclic LEFM Stress

σHoopLEFM = (9.45 psi)(81.00 in)/(0.070 in) = 10,935 psi

Dr. Patrick Safarian F&DT course notes provide that the hoop stress can be reduced by 15%

for the lower skin due to the presence of reinforcement (frames, bulkhead, floor beam and stringers)

σHoopLEFM = 10,935 psi (85%) = 9,295 psi

For both the condition (i) and condition (ii) stresses listed above, the maximum stress is shown and the

minimum stress is zero.

Torsional Loads

The pressurization stress is added to the shear stress in the skin due to fuselage torsion. The skin loads,

due to shear of the skin panels, are determined by considering the following:

* The skin panel is designed as a shear resistant web

* The skin buckles, due to shear, at a limit load * The skin shear is due to fuselage torsion from applied

side loads

* The skin panel is of a constant thickness, for local frame skin bays

* The skin panels (2024-T3) are considered to be of a uniform thickness of 0.040 inches throughout the

interior of the skin bay. Any doublers or chem milled thicknesses around the perimeter are neglected for shear of the panel. The frame spacing is 20.0 inches and the stringer spacing is 8.64 inches.

* The torsional loads are applicable only to the fatigue loading of the installation, the static strength analysis

does not consider the torsional loads.

a = 20.0 in b = 8.64 in t = 0.070 in a/b = 20.0/8.64 in = 2.31 KS = 5.6*(case 4) η

= 1.0 ** *Ref Figure 3.1.3A

**Elastic range only, no plasticity considered


DOA EASA.21J.560



Page 12 of 109

FC = (η)(K)(EC)(t/b)2 = (1.0)(5.6)(10.7E6 psi*)[(0.07 in/8.64 in)2] = 3,933 psi

*Ref 1, Table 3.2.4.0(e1), EC, 2024-T3 sheet

q = (3,933 psi)(0.070 in) = 275.31 lbs/in

The skin shear, critical for buckling, is considered to be a limit condition occurring at a limit 1.0g side load.

Normal Skin Shear = (275.31 lbs/in)/1.00 g = 275.31 lbs/in

Multiplying this value by a factor of 1.4 for large aircraft, accounts for the fluctuations in torsional stress that occur during a typical flight, compressing the full flight spectrum into a single cycle maximum stress.

τxy = (1.4)(3,933 psi) = 5,506 psi

This value is an ‘equivalent’ shear stress that incorporates the effects of a variable fuselage loading

spectrum into the single cycle maximum pressurization spectrum used in this analysis. This shear stress will be added to the independent lateral and longitudinal stress of the fuselage under 1-g load. NOTE:

due to the location of the repair, the tension stress due to the fuselage bending is not considered for

this analysis. However the tension stress due to the fuselage pressurization is considered. The shear and tensile stress are resolved into principal stresses to provide a tensile stress component with zero

shear.

σx = 5,468 psi σ y = 9,295 psi

σ1x,σ2x = σx/2 ± √[( σx/2)2 + τxy2] = 5,468 psi/2 ± √[(5,468 psi/2)2 + 5,506 psi2]

σ1x = 8,881 psi

σ2x = -3,413 psi


DOA EASA.21J.560



Page 13 of 109

σ1y,σ2y = σy/2 ± √[( σy/2)2 + τxy2] = 9,295 psi/2 ± √[(9,295 psi/2)2 + 5,506 psi2]

σ1y = 11,853 psi

σ2y = -2,558 psi

The angle at which the principal stress acts is given by

tan 2θPx = τxy/(σx/2)

θPx = (1/2)tan-1[τxy/(σx/2)] = (1/2)tan-1[5,506 psi/(5,468 psi/2)] = 31.8°

tan 2θPy = τxy/(σy/2)

θPy = (1/2)tan-1[τxy/(σy/2)] = (1/2)tan-1[5,506 psi/(9,295 psi/2)] = 24.92°

The following conservative value is used for the condition (i) fatigue loading conditions. σ1x = 8,881 psi

The LEFM stress previously calculated is updated to include the torsional effects. The longitudinal residual stress is considered to be recalculated. Note that the torsional stress is not considered in the hoop

condition (ii) stress.

σLongRes = 8,881 psi [(1.1(9.45 psi)+0.5 psi)/(9.45 psi)] = 10,239 psi

σHoopRes = 10,935 psi

σLongLEFM = 8,881 psi σHoopLEFM = 9,295 psi

Static Analysis

Antenna Inertia Load Condition

The maximum inertia load for the entire antenna installation is calculated below

Max inertia load = 6.5 lbs (9g) = 58.5 lbs

The maximum inertia load is not as critical as the side lift load condition. Therefore, analysis of the side lift load will also serve to substantiate the maximum inertia load condition.

Antenna Drag Load Condition

Drag Load = 14.83 lbs (x-dir)

*Ref. Fx, Section 3.0.3

The drag load is not as critical as the side lift load condition. Therefore, analysis of the side lift load will also serve to substantiate the maximum inertia load condition.

Antenna Side Lift Load Condition

The antenna is attached to the -11 hat section using ten ¼-28 screws into nutplates.

During the maximum operating condition the aerodynamic side lift load is considered applied at the c.g.

of the antenna and reacted in shear by the ten ¼-28 screws.

Side Lift Load = 440.87*lbs (y-dir)

*Ref. Fy, Section 3.0.3

Translation of the load up to the 10-32 screws produces an Mx moment. This moment is reacted as

bearing against the -11 hat section and in tension by a couple load between the inboard and outboard 10-32 screws.


DOA EASA.21J.560



Page 14 of 109

Py = Ry = 440.87 lbs (y-dir)

Mx Moment = (440.87 lbs)x(7.51-in) = 3,310.93 in-lbs

Rz = (3,310.93 in-lbs)/(1.64-in) = 2,018.86 lbs (z-dir)

Maximum Shear Load per screw = (440.87 lbs)/10 = 44.09 lbs (y-dir)

Maximum Tension Load per screw = (2,018.86 lbs)/5 = 403.77 lbs (y-dir)

M.S. (shear) = [(1,280*lbs)/(44.09 lbs x 1.15**) - 1] x 100% = LARGE

*Ref. ¼-in Threaded Fastener, Table 8.1.5(a), Ref. 1

**Ref. Fitting Factor


DOA EASA.21J.560



Page 15 of 109

M.S. (tension) = [(1,835*lbs)/(403.77 lbs x 1.15**) - 1] x 100% = +100%

*Ref. ¼-in Threaded Fastener, Table 8.1.5(b1), Ref. 1

**Ref. Fitting Factor

The entire couple load is transferred through the -11 hat section and reacted in tension and compression

by the two HL18-5 Hi-Loks attaching the -11 hat section to the -17 stringer support. Application of the

local z-directed loads on the -11 hat section produces local Mx bending moments. The width of the -11

hat section is modeled as a simple supported beam under two concentrated point loads.

Pz = 2,018.86 lbs (z-dir)

Rz = (2,018.86 lbs)x(1.64-in)/(6.22-in) = 532.3 lbs

Mx1 Bending Moment at Antenna Fasteners = (532.3 lbs)x(2.29*in) = 1,218.97 in-lbs

*Ref. Maximum distance from antenna fasteners to Hi-Loks

Mx2 Bending Moment at first rivet row = (532.3 lbs)x(0.86*in) = 457.78 in-lbs

*Ref. Maximum distance from Hi-loks to first rivet row

Tension Load per Hi-Lok = (532.3 lbs)/2 = 266.15 lbs

M.S.(tension) = [(1,940*lbs)/(266.15 lbs x 1.15**) – 1] x 100% = LARGE

*Ref. Tensile Strength of HL18-5 Hi-Lok, Figure 4.3D **Ref. Fitting Figure


DOA EASA.21J.560



Page 16 of 109

Check -11 hat section for bending:

Consider an effective bending width equal to the length of the antenna foot print, 12.43-in.

The -011 hat section is considered to support all out-of-plane loading in bending to the adjacent

stringers.

b = effective bending width = 12.43-in

t1 = combined thickness = 0.080-in + 0.070*in = 0.15-in

*Ref. Skin Thickness at Antenna fasteners

t2 = doubler thickness only = 0.080-in

σy1 = bending stress at Antenna = 6Mx1/(bt2) = 6 x (1,218.97 in-lbs)/[(12.43-in)x(0.15-in)2]

= 26,151 psi

σy2 = bending stress at outer rivet row = 6Mx’/(bt2) = 6 x (457.78 in-lbs)/[(12.43-in)x(0.08-in)2]

= 34,527 psi

M.S.(tension) = [(40,000*psi)/(34,527 psi) – 1] x 100% = +15%

*Ref. Fty, 0.063-0.128 2024-T3 Clad AL, Table 3.2.4.0(e1), Ref 1


DOA EASA.21J.560



Page 17 of 109

The local z-directed loads are transferred through the -17 stringer support and reacted in shear by the

two MS20470AD5 rivets attaching the -17 support to the -19 and -20 angle supports.

Pz = 532.3 lbs (z-dir)

Rz = (532.3 lbs)x(6.22-in)/(8.73-in) = 379.26 lbs (z-dir)

Z-dir Shear Load per rivet = (379.26 lbs)/(2) = 189.63 lbs

This shear load is later combined with a y-directed shear load. The -19 and -20 angle supports are fabricated as mirror opposites of each other, therefore analysis of the -19 angle support will also serve to substantiate

the -20 angle support.

The z-directed load is transferred through the -19 angle support and reacted in shear by the two MS20470AD6 rivets attaching the -19 angle to the existing 0.090-in thick 7075-T6 stringer.

Translation of the local z-directed load over to the AD6 rivets produces a local Mx moment. The local Mx

moments is reacted in shear by a couple load between the AD5 rivets attaching the -19 angle to the -17 support.


DOA EASA.21J.560



Page 18 of 109

Pz = Rz = 379.26*lbs

*Ref. Figure 4.3F

Mx Moment = (379.26 lbs)x(0.6-in) = 227.56 in-lbs

Ry = (227.56 in-lbs)/(0.75-in) = 303.41 lbs

Resultant Shear Load at -17 stringer supports = (303.41 2 + 189.632)0.5lbs = 357.79 lbs

M.S.(shear) = [(596*lbs)/(357.79 lbs x 1.15**) – 1] x 100% = +44%

*Ref. Shear Strength of AD5 rivet, Table 8.1.2(a), Ref. 1

**Ref. Fitting Figure

Translation of the z-directed loads over to the AD6 rivets in the stringers produces a local My moment which is reacted in shear by a couple load between the AD6 rivets.

Pz = 379.26 lbs

Rz1 = (379.26 lbs)x(0.6-in + 0.75-in)/(0.75-in) = 682.67 lbs

Rz = (379.26 lbs)x(0.6-in)/(0.75-in) = 303.41 lbs

M.S.(shear) = [(860*lbs)/(682.67 lbs x 1.15**) – 1] x 100% = +9%

*Ref. Shear Strength of D6 rivet, Table 8.1.2(a), Ref. 1

**Ref. Fitting Figure

σBR = P/A = (682.67 lbs)/[(0.094*in)(0.191 in)] = 38,023 psi

*Ref. Thickness of -19 angle

MS(bearing) = {(61,000 psi*/(38,023 psi)]-1} x 100% = +60%


DOA EASA.21J.560



Page 19 of 109

*Ref. Fbry, e/D=1.5, 'A' ≤0.249 2024-T3511 Alum Extrusion, Table 3.2.4.0(j 1), Ref. 1.

Therefore, the -11 hat section, -17 supports, -19 support angle, and -20 support angles are satisfactory

to react the maximum side lift load condition to the existing stringer structure.

Skin and Doubler Analysis

Analysis of the 1.31-in x 1.71-in hole treats it as any other holes through the skin. The doubler provides

an alternative load path for the forces in the skin. To show the doubler restores the strength to the

fuselage, the joint must pass the load through the doubler, which is a function of hole size. The doubler is conservatively considered 17.0 inches long and 5.35 inches wide.

Maximum lateral load to be transferred through the doubler;

= (19.85 psi*)(81.0 in**)(1.71-in) = 2,749.42 lbs

*Ref Section 3.0.1

**Ref fuselage radius

Maximum longitudinal load to be transferred through the doubler;

= (63,000 psi*)(0.070 in**)(1.31-in) = 5,777.1 lbs

*Ref 1, FTU, L, “B”, (.063-0.128), 2024-T3 clad sheet, Table 3.2.4.0(e1)

**Ref skin thickness

There are a minimum of fourteen NAS1097AD4 rivets considered effective on all sides of the doubler.

There are a minimum of forty-two NAS1097AD4 rivets considered effective on the fore and aft sides of

the doubler. These rivets are considered effective in transferring the load from the skin to the doubler. The doubler is 0.080 inches thick 2024-T3 aluminum and the skin is 0.070 inches thick 2024-T3

aluminum. There is no immediate ultimate strength allowable for the NAS1097AD rivet, therefore the

ultimate joint strength will be calculated by using a ratio of shear strengths. Per Ref. 1, (1/8), Table 8.1.2 (b), the Fsu value for a driven rivet fabricated from 7050-T731 aluminum alloy (E material) is 43

ksi. From the same table, the Fsu value for a driven rivet fabricated from 2117-T3 aluminum alloy (AD

material) is 30 ksi. Per Ref. 1, Table 8.1.2.2(n), the ultimate strength for an NAS1097E4 in 0.071 inch thick Clad 2024-T3 sheet is 497 pounds. Per Ref. 1, Table 8.1.2.2(n), the ultimate strength for an

NAS1097E4 in 0.063 inch thick Clad 2024-T3 sheet is 485 pounds.

Using interpolation between the two shear values of the sheets, the ultimate value for the NAS1097E4 rivet in the 0.070 inch thick skin is calculated below.

(497 lbs – 485 lbs)/(0.071 – 0.063)x(0.070 – 0.063) + 485 lbs = 495.5 lbs

Using the ratio between the two shear values of the rivets, the ultimate value for the NAS1097AD4 rivet in the 0.070 inch thick skin is calculated below.

Joint Ultimate Strength = (495.5 lbs)(30 ksi/43 ksi) = 345.7 lbs

Load per rivet (lateral load) = (2,749.42 lbs)/(14 rivets) = 196.4 lbs

Load per rivet (longitudinal load) = (5,777.1 lbs)/(42 rivets) = 137.55 lbs

M.S.(joint) = [(345.7 lbs)/(196.4 lbs x 1.15*)-1] x 100% = +53%


DOA EASA.21J.560



Page 20 of 109

*Ref. Fitting Factor

Checking the bearing allowable strength of the 0.0770 inch thick skin, reacting the lateral and

longitudinal rivet loads, the lateral loads are considered.

σBR = P/A = (196.4 lbs)/[(0.070 in)(0.129 in)] = 21,750 psi

MS(bearing)(ult) = [(101,000 psi*)/(21,750 psi)-1]100 = LARGE

*Ref. Fbru, e/D=1.5, 'A' (.063-.128 thk) 2024-T3 Clad Alum, Table 3.2.4.0(e1), Ref. 1.

MS(bearing)(lim) = {(70,000 psi*/(21,750 psi/1.5)]-1}100 = LARGE

*Ref. Fbry, e/D=1.5, 'A' (.063-.128 thk) 2024-T3 Clad Alum, Table 3.2.4.0(e1), Ref. 1.

The rivets are satisfactory to transfer the load from the skin to the doubler.

The hoop and longitudinal tension load is transferred from the skin of the aircraft through the rivets and into the doubler. The doubler is required to react the load. The tension load in the doubler is checked.

The effective width of the doubler for the feed through hole:

Doubler Effective Width (lateral load) = (17.0 – 1.71)in = 15.29-in

Stress = P/A = (2,749.42 lbs)/[(0.080 in)(15.29 in)] = 2,248 psi

Doubler Effective Width (longitudinal load) = (5.35 – 1.31)in = 4.04 in

Stress = P/A = (5,777.1 lbs)/[(0.080 in)(4.04 in)] = 17,875 psi

MS(Doubler-ult) = ((61,000 psi*/(3**(17,875 psi))-1)100 = +13%

*Ref. Ftu, (0.063 – 0.128 thk) 2024-T3 Clad Alum, Table 3.2.4.0(e1), Ref. 1.

**Ref 12, stress concentration factor for a tension loaded plate, round hole

MS(Doubler-Lim) = ((40,000 psi*/(3**(17,875 psi/1.5))-1)100 = +11%

*Ref. Fty, (.063-.128 thk) 2024-T3 Clad Alum, Table 3.2.4.0(e1), Ref. 1.

**Ref 12, stress concentration factor for a tension loaded plate, round hole

Alternatively consider the ultimate pressurization load scenario in which the hoop stress and longitudinal pressurization stress act simultaneously at the center feed through hole. The rivets attaching the doubler

to the skin are considered to equalize the stress in the skin and doubler.

σx (ult) = (19.85 psi)(81.0 in)/[(2)(0.070 in)] = 11,485 psi

σy (ult) = (19.85 psi)(81.0 in)/(0.070 in) = 22,969 psi

Skin Remote Longitudinal Tension = (11,485 psi)(0.070 in) = 803.95 lbs/in

Skin Remote Hoop Tension = (22,969 psi)(0.070 in) = 1,607.83 lbs/in

The maximum load to be transferred to the doubler from the skin is a function of rivet joint strength, and

rivet pitch. Consider two rows of rivets effective in transferring the load. The maximum allowable load per

rivet was previously shown as 345.7 pounds, and there are two rows of rivets with a maximum 0.9 inch rivet pitch. The maximum skin stress at the feed through hole is calculated below.

Skin Longitudinal Tension = (803.95 lbs/in) - 2(345.7 lbs)/(0.9 in) = 35.73 lbs/in

Skin Hoop Tension = (1,607.83 lbs/in) - 2(345.7 lbs)/(0.9 in) = 839.61 lbs/in

Skin Longitudinal Stress = (35.73 lbs/in)/(0.070 in) = 510 psi

Skin Hoop Stress = (839.61 lbs/in)/(0.070 in) = 11,994 psi

As the stress in the doubler cannot exceed that of the skin, a calculation of the doubler and skin load is

performed considering equal stress in each element.


DOA EASA.21J.560



Page 21 of 109

Equalized Longitudinal Stress = (803.95 lbs/in)/(0.070 in + 0.080 in) = 5,360 psi

Equalized Hoop Stress = (1,607.83 lbs/in)/(0.070 in + 0.080 in) = 10,719 psi

The maximum stress in the skin, at the feed through hole, is considered for the biaxial tension load condition, using the larger skin stress as calculated in the two scenarios above.

Maximum Skin Longitudinal Stress = 5,360 psi = σ2

Maximum Skin Hoop Stress = 11,994 psi = σ1

The stress concentration factor for an ovaloid in an infinite width thin sheet biaxially stressed is used to determine the stress values at the longitudinal and lateral edges of the feed through hole

(Ref 12, Chart 4.62b, infinite thin sheet, ovaloid hole in biaxial tension).

a = 1.71-in/2 = 0.855

b = 0.655-in

r = 0.66-in

r/a = 0.66/0.855 = 0.772

for σ2 = σ1/2

Kt = 2.79

σmax = Ktσ1 = (2.79)x(11,994 psi) = 33,463 psi

MS(skin) = [(61,000 psi*/33,463 psi)-1]100 = +82%

*Ref 1, Ftu, Table 3.2.3.0(e1), 'A', 0.063-0.128, 2024-T3 clad sht

The doubler and fasteners are satisfactory to react to the ultimate cabin pressurization loads.


DOA EASA.21J.560



Page 22 of 109

The internal pressure load is applied to the antenna over the 1.31 x 1.71-in feed through hole.

A = (π)(d/2)2 = (π)(0.66 in/2)2 + 0.39-in x 1.31-in = 0.85 in2 F = (ΔPult)(A) = (19.85 psi)(0.85 in2) = 16.87 lbs

MS(tension) = [(1,835 lbs*)/(16.87 lbs) -1] 100 = LARGE

*Ref. Allowable, ¼-28 threaded fastener, Table 8.1.5(b1), Ref. 1

The fasteners and doubler are satisfactory to react the applied loads from the pressurization of the aircraft.

Fatigue Assessment and Damage Tolerance Assessment

The capabilities of the fuselage alterations are analyzed to ensure that the structure could tolerate serious fatigue, corrosion, or accidental damage during the operational life of the aircraft.

Rogue Flaw, Geometry

The models for the rogue flaw serve to establish the threshold inspection interval based upon the

chosen inspection type. The rogue flaw is conservatively grown to the first link up only provided failure

does not occur prior to the first linkup. The initial crack scenario considers one 0.05 inch corner crack with an opposing 0.010 inch corner crack at a hole.


DOA EASA.21J.560



Page 23 of 109

Rogue Flaw, Crack Growth Methodology

The 0.010 inch and 0.05 inch initial part through cracks are grown using the AFGROW advanced modeler. The initial crack sizes are determined by the location and type of flaw.

The initial part through cracks have an "a" thickness dimension equal to the thickness of the skin and

a "c" length dimension of either 0.010 inches or 0.05 inches as shown in the figures above. The plate

width is considered as four times the rivet pitch minus one rivet diameter.

Prior to linkup, the model with a 0.010 inch crack and a 0.05 inch crack at the center hole is analyzed using the calculated tensile, bending, and bearing stresses to be applied at each hole. The lead cracks

are grown in a second model using the two holes adjacent to the center hole using the same loading.


DOA EASA.21J.560



Page 24 of 109

The term "linkup" refers to the scenario where the edges of the crack emanating from the center hole

propagate to the point which they reach the edge of the adjacent holes.

Crack Growth Properties

The crack growth rate analyses were calculated using the Air Force Research Laboratory software

package AFGROW, using the NASGRO equation. The stress intensity factor expressions for a pin loaded

hole (rivet hole) and for a crack emanating from a tension and bending loaded hole are available as

analysis options within the AFGROW computer program. The AFGROW material database for 2024-T3 T-L clad sheet, shown in Figure 5.0.3A, was employed in the analysis. This resulted in the crack growth-

rate data shown in Figure 5.0.3B.

The da/dN data is coded inside the AFGROW program based upon the NASGRO crack model, Refer

Anon’ AFRL-VA-WP-TR-1999-3016 AFGROW Users Guide and Technical Manual Air Vehicle Directorate, Air Force Research Laboratory.


DOA EASA.21J.560



Page 25 of 109

NASGRO (Forman, Newman, de Koning and Henriksen) Equation

The elements of the NASGRO (Version 3.00) crack growth rate equation were developed by Forman

and Newman at NASA, de Koning at NLR, and Henriksen at ESA. It has been implemented in AFGROW as follows:

Where C, n, p, and q are empirically derived, and

The coefficients are:

Here, 'a' is the plane stress/strain constraint factor, and Smax/σo is the ratio of the maximum applied

stress to the flow stress. These values are provided by the NASGRO material database for each material.

Where:

ΔKo - threshold stress intensity range at R=0

a - crack length (a or c in AFGROW) ao - intrinsic crack length (0.0015 inches or 0.0000381 meters)

Cth - threshold coefficient

The values for ΔKo and Cth are provided by the NASGRO material database for each material. The NASGRO

equation accounts for thickness effects by the use of the critical stress intensity factor, Kcrit.

Where:

KIc - plane strain fracture toughness (Mode I)

Ak - Fit Parameter Bk - Fit Parameter

t – Thickness

to - reference thickness (plane strain condition)


DOA EASA.21J.560



Page 26 of 109

The plane strain condition is:

The values for KIc, Ak, and Bk are provided by the NASGRO material database for each material. Although the plane strain thickness, t0, is defined by the equation shown above, Kcrit will asymptotically approach

KIc as the actual thickness gets larger than t0.

Threshold Inspection Interval

The Threshold Inspection is determined from the smaller cycle count of; (Nfinal/2), (Ndetectable), or

1/2 the OEM established life limit of the component. If no OEM established life limit exists for the component a design service goal of 20,000 cycles is considered.

For the inspection purposes of this analysis, one cycle is equivalent to one flight.

The number of cycles (Nfinal) is determined by the crack growth analysis at which the detail is considered to reach a critical crack length.

The number of cycles (Ndetectable) is determined by the crack growth analysis at which a crack is

considered to reach a detectable or inspectable crack size. The detectable flaw size considered is a

function of the crack type and of the type of inspection proposed for the structure as shown in the Table 5.0.4A.


DOA EASA.21J.560



Page 27 of 109


DOA EASA.21J.560



Page 28 of 109

Recurring Inspection Interval

The Recurring or Subsequent Inspection interval is determined from either; 1/2 the threshold inspection

interval, or 1/3 of (Nfinal - Ndetectable).

When considering Multiple Site Damage (MSD) with Wide Spread Fatigue Damage (WSFD) the Recurring

or Subsequent Inspection interval is determined from either; 1/2 the threshold inspection interval, or

1/3(Nfinal).

The recurring inspection interval is paired with the threshold inspection for each respective case.

Multiple Site Damage

Multiple Site Damage (MSD) considerations are evaluated for those areas of the installation that are coincident with existing or replaced fasteners of which every hole is considered to have a 0.05 inch long

crack at each edge growing until a net section failure occurs, with the assumption that there is an infinite

number of cracked holes. The MSD evaluation satisfies the requirements of 14 CFR 25.571(b) "Damage at multiple sites due to prior fatigue exposure must be included where the design is such that this type

of damage can be expected to occur".

For the Wide Spread Fatigue Damage (WSFD) considered in the analysis, the critical crack length is based solely on the Net Section Yield (NSY) failure criteria rather than the Linear Elastic Fracture

Mechanics (LEFM) criteria.

The critical crack lengths for the MSD models are determined using 90% of the skin material's 'B' basis

ultimate tensile strength (Ftu). The 90% factor is a correction knockdown factor to account for reductions

of the jointed sheet.

Where:

Rp ~ rivet pitch

Ftu ~ skin ultimate tensile strength, 'B' basis

σLongRes ~ longitudinal residual strength required

σHoopRes ~ hoop residual strength required

D ~ effective hole diameter

MSD Critical Crack Length (longitudinal)

= {Rp/[(2)(Ftu)(0.9)]}[(Ftu)(0.9) - σLongRes - (D/Rp)(Ftu)(0.9)]

MSD Critical Crack Length (hoop)

= {Rp/[(2)(Ftu)(0.9)]}[(Ftu)(0.9) - σHoopRes - (D/Rp)(Ftu)(0.9)]

Fastener Hole Geometric Factors The geometric factors;


DOA EASA.21J.560



Page 29 of 109

The geometric factors; pAF (hole with opposing part through cracks subject to bending, bearing,

and tension) and pMSD (infinite series of collinear cracks), are used to create the stress intensity

factor for the MSD analysis. The following stress intensity expression is considered:

K = (PMSD)(PAF)(Oref)V[(n)(c)]

The AFGROW software program is used for the MSD analysis considering the following:

oref ~ AFGROW reference

stress c ~ crack length

The geometric factors for the holes with MSD subject to bending, bearing, and tension are

calculated using AFGROW. The specific reference stress, tension stress ratio, bearing stress ratio,

and bending stress ratio are the same as those calculated for the rogue flaw model. A sheet or

plate width of TCD =295 inches is considered for a simple model (oblique through crack) with a

central hole having opposing double corner cracks of 0.05 inches. The AFGROW program is used

to generate an Excel spreadsheet containing the crack length and corresponding pAF values. The

data in the spreadsheet is then curve fit to define the pAF as a function of crack length, which is

compounded with a crack interaction function to create the stress intensity factor (K).

The determination of the geometric factor for an infinite series of collinear cracks (PMSD), is obtained

using the following equation based upon the methodology of Murakami's Stress Intensity

Handbook.

A simplified analysis for crack growth at multiple collinear holes can be simulated by analyzing the

propagation of two cracks at a single central hole in a sheet of finite width equal to the rivet pitch.

This second simplified approach is employed in this analysis where the installation is coincident

with existing holes in the airframe structure (doubler fasteners common to stringers or frames),

where the MSD evaluation is required.

When considering MSD the threshold inspection is determined by either the rogue flaw or fatigue

methods, however the recurring inspection interval is the lesser of the rogue flaw, fatigue or the

following where (Ncrtical) is the number of cycles required to grow the crack to the critical length

and (Ndetectable) is the number of cycle required to grow the crack to an (inspection method

dependent) detectable length

Recurring Inspection Interval (MSD) = (Ncrtical - Ndetectable)/3

Loading Spectrum

The longitudinal loading spectrum is defined as the repeated application of the normal operating

differential pressure in addition to the bending stress calculated for fuselage bending. The crack

growth analysis is performed using a constant amplitude stress spectrum and R ratio of 0.0

\ (Rp ) ta

n

n ( c + j )

P M S D ~ R p

U { C + l )


DOA EASA.21J.560



Page 30 of 109

(minimum stress / maximum stress). For crack growth analysis, one pass of the spectrum is

considered the equivalent of one flight cycle.

The lateral loading spectrum is defined as the repeated application of the normal operating

differential pressure and external aerodynamic pressures. The crack growth analysis is performed

using a constant amplitude stress spectrum and R ratio of 0.0 (minimum stress / maximum stress).

For crack growth analysis, one pass of the loading spectrum is considered the equivalent of one

flight cycle.

The longitudinal loading spectrum is defined as the repeated application of the bending stress

calculated for fuselage bending.

Condition (i) Fuselage Loading Due to Bending and Pressure - Lateral Cracking

The critical areas for crack initiation and growth are between rivets in the outer row of rivets and

cracks emanating from the feed through hole.

The analysis considers a crack that grows from the edge of one rivet to the edge of the adjacent

rivet in the same row. These critical crack locations are illustrated below in Figure 5.1A. Only the

skin is checked as the skin has a higher stress than the doubler, the inspection of the doubler shall

follow the same intervals and techniques as the skin.

Joint Strength

The fatigue analysis of the reinforced skin is based on the Neutral Line Analysis (NLA), Reference

Schijve, et al [9,19], described in the appendix A of this report. This method is performed in two

steps. The first is to calculate the fastener forces in the rivet rows used to attach the doubler to the

skin. The second is to calculate the secondary bending of the skin- doubler system using the method

presented by de Rijck, et al [18]. Swift’s fastener flexibility model [20] is employed in the analysis.

The geometry for this skin-doubler model is shown in Figure 5.1B.


DOA EASA.21J.560



Page 31 of 109

The first four rivet rows on the LH side and the last four rivet rows on the RH side are modeled in

the Neutral Line Model. NAS1097AD4 rivets are installed in each of these rows with a rivet row

spacing and rivet pitch as shown in the above Figure 5.1B.

Skin tension = (0.070-in)(8,881 psi*) = 621.67

lbs/in *Ref: Section 3.1.3, Longitudinal LEFM stress

The analysis of the riveted joints uses the displacement-compatibility model shown in Figure 5. IB,

with Swift’s fastener flexibility.

The stiffness of the skin and doubler are calculated using the formula

where t represents the thickness, 0.070-in and 0.080 in for the skin and doubler, respectively; W

representes the constant width of the douber, W = 5.19 in (Conservative Consideration); and the element

length L is the distance between rivet rows. This element length is also assigned to the “skin” elements to

the left between points 0 and SL3 and between points SR3 and N.

The compliance for Swift’s fastener model is defined for aluminum rivets as


DOA EASA.21J.560



Page 32 of 109

where D represents the fastener diameter, ti and t2 represent the skin and

doubler thickesses, defined above, and E = 10.5 (10)6 psi (Ref. Table 3.2.4.0(e1) MMPDS-06, represents

Young’s modulus for the skin, doubler and fastener. The resulting stiffness for a row of fasteners is

calculated as

where Nf represents the number of fasteners in the row, thus N f = 4 and N f = 4 for the first and

fourth forward rows of rivets repectively.

The applied load P , is defined in terms of the longitudinal stress as

where the minus sign denotes the fact that the force acts in the -x direction. This load is used to

calculate displacements are calculated from the assembled stiffness matrix.

o

The maximun displacement occurs at point 0 and was calculated as u 0 = -9.8949 x 10- in. The resulting

fastener loads are tabulated as

The greatest bearing stress is calculated as

brg = 140.08 lb / [(0.070-in)(0.163 in*)] = 12,277 psi

*Ref: Consider a hole size of 0.163-inch for the 1/8 rivet in a 0.129-in hole which has a head diameter of 0.196-in.

The corresponding bearing stress ratio kb is calculated as

kbrg = 12,277 psi / 8,881 psi = 1.382

The skin and doubler Stress resultants are reported in Tables 5.1B and 5.1C.


DOA EASA.21J.560



Page 33 of 109

The secondary bending stresses generated by the doubler were analyzed using the Neutral Line Method

suggested by Schijve [9,19]. This particular analysis incorporates the effects of fastener flexibility as

described by de Rijck [18].

The corrective moments were computed as described in References 11 & 12 and are summarized in Table

5.1D.

Table 5.1D - Corrective moments

Rivet Moment (in-lb)

AM1L -66.022

AM2L 33.048

AM3L 18.071

AM4L 11.302

AM4R -11.302

AM3R -18.071

AM2R -33.048

AMIR 66.022


DOA EASA.21J.560



Page 34 of 109

The neutral line displacement solution is presented in Figure 5.1E, where the units of both axes are

inches.

The bending stress is calculated from the differential moment-deflection equation.

which can be written in terms of the homogeneous portion of the neutral line solution as

M = E

Substituting this moment into the bending stress equation for a flat plate

The corresponding bending stress ratio k b is calculated as

k b = 5,620 psi / 8,881 psi = 0.633

The bypass stress is calculated as

The corresponding bypass stress ratio k b y p a s s is calculated as

k b y p a s s = 6,568 psi / 8,881 psi = 0.740


DOA EASA.21J.560



Page 35 of 109

Fatigue Life

The fatigue life is considered using three methods and the most conservative result is utilized as the fatigue

based threshold.

Method 1

Method 1 is the 'Existing Life Limit' approach, which considers existing OEM life limitations placed on the

detail, assembly, or installation.

No life limits are listed in the applicable Airworthiness Limitations sections of the approved aircraft manual.

However, LOV of the 757-200 airplane is 50,000 FC per 14 CFR Paragraph 121.1115(f) amendment 25-

360. Therefore the Design Service Goal (DSG) of 50,000 cycles is considered, in the fatigue analysis.

DSG (1) = 50,000 cycles

Threshold Inspection (1) = 50,000 cycles/2 = 25,000 cycles Recurring Inspection

(1) = 25,000 cycles/2 = 12,500 cycles

Method 2

By applying the following series of managed scatter factors as shown in Scatter Reduction Factor*

methods A and B that represents the alteration/installation on the aircraft, the fatigue life of the alteration

is determined. *Reference Patrick Safarian, DTA Seminar, Lesson 19, Spirit Aviation, Inc., Wichita, KS,

Feb 25-27, 2013.

Scatter Reduction Factor, A

To manage scatter use the following four factors:

Testing Factor 0.7 < F < 1.0*; To account for differences in scale and fidelity of the test, including the

extent to which the loading of the test article represents the actual structure.

Confidence Factor Use 0.7 for 95% lower confidence bound*; A statistical factor to address the

uncertainty in the final value caused by the limited sample size.

Reliability Factor* 0.48/Alum, 0.26/Steel*; A conversion factor to obtain a reliable life value from mean

or characteristic life data.

*(Ref 9, Scatter is less at high stress amplitudes, and larger at low stress amplitudes)

Scale Factor 0.33 < F < 0.50*; A factor to adjust the design life value based on the percentage of details

in the specimen to the number of detail in the actual structure.

*see Appendix C, recommended factors and Tables for Reliability and Scales factors.

Scatter Reduction Factor = Testing Factor x Confidence Factor x Reliability Factor x Scale Factor


DOA EASA.21J.560



Page 36 of 109

Note, the life of fatigue life of the subject is reduced by multiplying the MMPDS fatigue life by this Scatter

Reduction Factor.

Scatter Reduction Factor, B

To manage scatter use the following three factors:

Scale Factor 1.0 < F < 2.0*; To account for differences in scale and fidelity of the test, including the


Load Factor 1.0/Spectrum, 1.5/Const Amp Loading*; To account for the effects of loading type on the

fidelity of the data

Reliability Factor 2.75/alum, 3.5/Steel; A conversion factor to obtain a reliable life value from mean or

characteristic life data.

* see Appendix D, recommended factors and Tables for Reliability and Scales factors.

Scatter Reduction Factor = Scale Factor x Load Factor x Reliability Factor

Note, the life of fatigue life of the subject is reduced by dividing the MMPDS fatigue life by this Scatter

Reduction Factor.

Method 2 is the 'Fokker empirical prediction method' (Reference 9, Chapter 18). The stress concentration

factors for tension, bearing, and bending are calculated for the loaded rivet hole. Predictions are

extrapolated from this curve by accounting of three contributions to the stress concentration at the rivet

holes of the critical end row. The contributions are associated with (i) load transmission by the rivet (pin

loading on hole), (ii) bypass loading of the rivet rows, and (iii) increased stress by secondary bending. The

equation used is the following:

In this equation y is the percentage of the load transmitted to the other sheet in the critical row. Then, (1-

y) is the percentage of the bypass load. The factor k is the secondary bending factor.

Y = (Rivet Load/Rivet Pitch)/(Skin Tension)

= (140.08 lbs/0.90-in)/(621.67 lbs/in) = 0.251

As described in Peterson, Ref. 12. The case of a pinned joint in an infinite thin element has been solved

mathematically by Bickley (1928). The finite-width case has been solved by Knight (1935), where the

element width is equal to twice the hole diameter d and by Theocaris (1956) for d/H = 0.2 to 0.5.

Experimental results (strain gage or photoelastic) have been obtained by Coker and Filon (1931),

Schaechterle (1934), Frocht and Hill (1940), Jessop, Snell and Holister (1958), and Cox and Brown (1964).

(See Peterson for descriptions of these references).


DOA EASA.21J.560



Page 37 of 109

In Figure 5.1E, (Chart 4.67) the K t n b curve corresponds the Theocaris (1956) data for d/H = 0.2 to 0.5.

The values of Frocht and Hill (1940) and Cox and Brown (1964) are in good agreement with Chart 4.67,

although slightly lower. From d/H = 0.5 to 0.75 the foregoing 0.2-0.5 curve is extended to be consistent

with the Frocht and Hill values. The resulting curve is for joints where c/H is 1.0 or greater. For c/H = 0.5,

the Ktn values are somewhat higher.

From Eq. (4.85), Ktnd = Ktnb at the d/H = 'A It would seem more logical to use the lower (full line) branches

of the curves in Figure 5.1G (Chart 4.67), since, in practice; d/H is usually less than 'A This means that Eq.

(4.84), base on the bearing area, is generally used.

The rivets are considered countersunk in the doubler. This analysis will consider an initial through hole of

the average skin hole of 0.163 inches.

The geometry concentration factor for the pin (Kt, pin) (i) is determined from the following graph, Peterson Chart

4.67:


DOA EASA.21J.560



Page 38 of 109

This data results in following “net” stress concentration factor. Ktn, pin =1.30

The geometry concentration factor for the hole in tension (Ktg, hole) (ii) is determined from Peterson Chart

4.1:


DOA EASA.21J.560



Page 39 of 109

This data results in the following “net” stress concentration factor. Kta= 2.56

The geometry concentration factory for the hole in bending (Ktg, bending) (iii) is determined from Peterson

Chart 4.83. The calculations are considered for simple bending.

Figure 5.1G - Peterson Chart 4.1, Kt Hole in Tension


DOA EASA.21J.560



Page 40 of 109

Ktn, hole bending = 1.6


DOA EASA.21J.560



Page 41 of 109

The ratio of the bending stress to the applied tension stress (k) (iii) is obtained:

kb=0.633*

*Ref. below Figure 5.1D

The total stress concentration factor due to pin load, bypass, and secondary bending is denoted:

Kt= (γ)(Kt, pin) + (1- γ)(Kt, hole, tension)+ (kb)(Kt, hole bending)

Ktn= (0.251)(1.30) + (1–0.251)(2.56) + (0.633)(1.68) =3.307

The Ktn value that is calculated is input into the S/N data curve presented in the MMPDS Ref. 1 to determine the life of the installation. The life of the installation is calculated using a logarithmic interpolation.

Effective Stress in Kt= 2 panel: 8,881psi

Seq= Smax(1- R)0.68 = 8.881ksi(1- 0)0.68 = 8.881ksi

log(Nf) = 9.2- 3.30[log(Seq– 8.5)] = 9.2- 3.30[log(8.881– 12.3)] = NONREAL

Nf= NEAR INFINITE ≈ 1010


DOA EASA.21J.560



Page 42 of 109

Effective Stress in Kt = 4 panel: 8,881

psi

Seq = Smax(1-R)0.66 = 8.881 ksi(1-0)0.66 = 8.881 ksi

log(Nf) = 8.3 - 3.30[log(Seq – 8.5)] = 8.3 - 3.30[log(8.881 – 8.5)] = 9.682

Nf = 10(9.682)

The fatigue life corresponds to the Kt of 3.307 is calculated using a logarithmic interpolation.

Fatigue Life = 10(Nf) = 10(9.791 ) = 6.18 x 109 cycles

With Scatter Reduction Factor A

Fatigue life = (6.18 x 109 cycles) [(0.7a)(0.7b)(0.48c)(0.38d)] = 552,343,680 cycles aRef. Testing Factor bRef. Confidence Factor


DOA EASA.21J.560



Page 43 of 109

cRef. Reliability Factor dRef. Scale Factor

With Scatter Reduction Factor B

Fatigue life = (6.18 x 109 cycles)/ [(1.5a)(1.5b)(2.75c)] = 998,787,878 cycles aRef. Scale Factor bRef. Load Factor cRef. Reliability Factor

The recurrent inspection interval for Part 25 aircraft are not to be based on fatigue life per Patrick Safarian DTA

Seminar, Feb 25-27, 2013. Fatigue life calculated with both A and B scatter factors are acceptable. For this

analysis the fatigue life with scatter factor of A is considered.

The acceptable conservative fatigue life limit for the longitudinal methods are therefore, based on Method 1.

Fatigue Life = 50,000 cycles

Threshold Inspection = 50,000 cycles/2 = 25,000 cycles

Recurring Inspection = 25,000 cycles/2 = 12,500 cycles

Crack Growth Analysis for Crack Between Rivets in Forward/Aft Row - Rogue Flaw

For the rogue flaw damage model the residual strength is based upon the combination of KMAX and NSY.

The simplified model for the lateral skin crack between two adjacent rivets in the side row of the skin is modeled

as a pin loaded hole in a flat sheet. The sheet width is taken to be twice the width of the skin bay. The applied

stress is the reference stress. The reference stress and applicable stress fractions are calculated at the beginning

of this section.

Effective skin width = 17.28 in


DOA EASA.21J.560



Page 44 of 109

Model #1 Fatigue Loading


DOA EASA.21J.560



Page 45 of 109

The residual strength curve is constructed by plotting the results of the AFGROW crack growth model

for the left 0.05 inch long initial crack. The 1000-9000 series aluminum, 7076-T6 Al, [Clad; plt & sht;

L-T & T-L; LA ] material properties from the AFGROW data base are used for the calculations.

E = 10,400 ksi

υ = 0.33 F

ty = 75 ksi,

KIC = 27 ksi√in

KC = 54 ksi√in

The applied fatigue remote stress and required residual strength are constant. The NSY curve is

defined by:

The Net Section Strength is determined by iteration (using a 0.001 convergence tolerance). The βC

factor, cycle count, and crack length for each crack growth increment are obtained from the AFGROW

output data. The critical K value (apparent fracture toughness) is defined by:

Where: index ~ Stress State Index (6 – plane strain, 2 – plane stress) The “Allowable Stress” for KCrit

is calculated as follows:

The first intersection of the 'Residual Strength Required' line with either the KMAX or NSY lines indicates

a critical crack length and resulting failure of the detail.

The chart on the following page illustrates the residual strength characteristics of the model considering

the Net Section Yield (NSY) and KMAX failure criterion.


DOA EASA.21J.560



Page 46 of 109

Evaluation of the residual strength chart indicates that the right 0.05 inch initial crack has reached a

critical crack length of 0.71 inches base on Kmax criteria. Note that the non-linear nature of the NSY curve is due to the addition of the yield zone size to the crack length at the crack tips.

The AFGROW crack growth curve for the 0.05 inch initial crack is shown on the next page. Note that the

chart shown is for a crack growth model which propagates the crack to the next free edge, in the

advanced model case when a crack reaches an adjacent hole.

The AFGROW program also has options to halt the crack growth at a user specified crack length which

is used to determine the number of cycles required to reach a detectable crack length, and option to

halt the program for NSY and KMAX failure criteria.

The first run of the program halts the crack growth at the user specified detectable crack length (with

the option selected to halt at a critical crack length). The second run of the program, using the same

model, halts the program when either the NSY or KMAX criteria have been exceeded. A third run of the program is conducted which allows the cracks to reach a free edge for illustrative purposes.


DOA EASA.21J.560



Page 47 of 109

The results of the crack growth analysis for a crack growing from one rivet hole to the adjacent rivet in

the outer row are shown on the next two pages. The analysis predicts 62,223 cycles are required to

grow the crack to an inspectable length, (0.196-in – 0.129-in)/2 + 0.10 in = 0.1335.

The inspections are done from the outside of the airplane and has direct access to the skin. Therefore,

the HFEC (High Frequency Eddy Current) method can be used.


DOA EASA.21J.560



Page 48 of 109

The analysis predicts that 179,358 cycles are required to grow the crack to a length of 0.71 inches which

is the critical crack length.


DOA EASA.21J.560



Page 49 of 109

Threshold Inspection (1) = 62,223 cycles

Threshold Inspection (2) = 179,358 cycles / 2 = 89,679 cycles

Recurring Inspection (1) = (179,358 cycles – 62,223 cycles) / 3 = 39,045 cycles

Recurring Inspection (2) = 89,679 cycles / 2 = 44,839 cycles

The recommended threshold inspection and recurring inspection interval (for the longitudinal load case)

are:

Threshold Inspection = 62,000 cycles

Recurring Inspection = 39,000 cycles

The following data is extracted from the AFGROW output file:

AFGROW 5.2.2.18 2/10/2015 17:07

**English Units [ Length(in), Stress(Ksi), Temperature(F) ]

Crack Growth Model and Spectrum Information

Title: Longitudinal Loading, Lateral Cracks, 1st Rivet Row

Load: Axial Stress Fraction: 0.74, Bending Stress Fraction: 0.633, Bearing Stress Fraction:


DOA EASA.21J.560



Page 50 of 109

1.382

Advanced Models Thickness : 0.070 Width : 17.280

Crack #1 (Corner Crack at Hole) Length = 0.05 Position: Hole Left

Crack #2 (Corner Crack at Hole) Length = 0.01 Position: Hole Right

Hole #1 (Hole) Diameter = 0.163 Offset = 7.74



Young's Modulus =10600 Poisson's Ratio =0.33

Coeff. of Thermal Expan. =1.29e-005

No crack growth retardation is being considered

Determine Stress State automatically (2 = Plane

stress, 6 = Plane strain) No K-Solution Filters

The Forman-Newman-de Koning- Henriksen (NASGRO) crack growth relation is being used For Reff < 0.0, Delta K = Kmax Material: 1000-9000 series aluminum, 2024-T3 Al, [ Clad; plt & sht; T-L ]

Plane strain fracture toughness: 29 Plane stress fracture toughness: 58 Effective fracture toughness for surface/elliptically shaped crack: 41 Fit parameters (KC versus Thickness Equation): Ak= 1, Bk=1 Yield stress: 48 Lower 'R' value boundary: -0.3


DOA EASA.21J.560



Page 51 of 109

Upper 'R' value boundary: 0.7 Exponents in NASGRO Equation: n=2.601, p=0.5, q=1 Paris crack growth rate constant: 2.44e-008 Threshold stress intensity factor range at R = 0: 2.9 Threshold coefficient: 1.5 Plane stress/strain constraint factor: 1.5 Ratio of the maximum applied stress to the flow stress: 0.3

Failure is based on the current load in the applied spectrum Cycle by cycle beta and spectrum calculation

**Spectrum Information

Constant amplitude loading Spectrum multiplication factor: 8.881 SPL: 0 The spectrum will be repeated up to 10000000 times Total Cycles: 1 Levels: 1 Subspectra: 1 Max Value: 1 Min Value: 0

No Spectrum Filters

Stress State in 'C' direction (PSC): 2

Transition will be based on K max or 95% thickness penetration Criteria Length Beta Tension Beta

Compression R(k) R(final) Delta-K D( )/DN

Crack #1 Left Tip C 0.05 2.1893 2.1893

0.0000 0.0000

7.7061e+000 1.2199e-006

Right Tip C

0.01 4.0090 4.0090 0.0000 0.0000

6.3106e+000 6.7272e-007 Left Tip A 0.07 2.0152 1.2055 0.0000

0.0000 8.3926e+000 1.5793e-006

Right Tip A

0.07 0.4778 0.4778 0.0000 0.0000

1.9897e+000 0.0000e+000 Max stress

8.881, r = 0.00, 0 Cycles,

Constant amp.: 1, Pass: 1

*********Transition at 95% thickness penetration

Length Beta Tension Beta Compression R(k) R(final) Delta-K D( )/DN

Crack #1 Left Tip C 0.05 2.0449 2.0449

0.0000 0.0000

7.1977e+000 9.8995e-007

Right Tip C

0.01 3.8987 3.8987 0.0000 0.0000

6.1370e+000 6.1688e-007 Max stress

8.881, r = 0.00, 0 Cycles,

Constant amp.: 1, Pass: 1


Crack #1 Left Tip C 0.727637343.4764 7343.4764 0.0000 0.0000 9.8604e+004

1.0000e-001 Right Tip C

0.66274 1.4853 1.4853

0.0000 0.0000 1.9034e+001 1.9219e-005 Max stress

8.881, r = 0.00, 179453 Cycles, Constant amp.

: 179454,

Pass: 179454


Crack #1 Left Tip C 0.727637343.4764 7343.4764 0.0000 0.0000 9.8604e+004

1.0000e-001 Right Tip C

0.66274 1.4853 1.4853

0.0000 0.0000 1.9034e+001 1.9219e-005 Max stress

8.881, r = 0.00, 179453 Cycles, Constant amp.

: 179454,

Pass: 179454 ++++++Cycles 179454 Crack 1, Edge 1 touched 2 .................»Crack [0] Dim[0] transitioned to a hole [0]


DOA EASA.21J.560



Page 52 of 109

Fracture has occurred - run time: 0 hour(s) 0 minute(s) 13 second(s)

Crack Growth Analysis for Crack Extending from edge of feedthru - Rogue Flaw

For the rogue flaw damage model the residual strength is based upon the combination of KMAX and

NSY.

The simplified model for the lateral skin crack between the feed through hole and the two adjacent

rivets in the skin is modeled as an open hole in a flat sheet. The sheet width is taken to be twice the width of a skin bay, 17.28-in.

Applied stress to the skin at the feedthru hole = (299.36*lbs/in)/(0.070**in) = 4,277 psi *Ref

Table 5.1B **Ref: skin thickness

Residual Stress = (10,239*psi)x(4,277 psi)/(8,881 **psi) = 4,931 psi

*Ref: Longitudinal residual stress, calculated in Section Condition (i) Far Field Stress (Longitudinal

Direction)

**Ref: Longitudinal LEFM stress, calculated in Section Condition (i) Far Field Stress (Longitudinal

Direction)



DOA EASA.21J.560



Page 53 of 109



DOA EASA.21J.560



Page 54 of 109

The results of the crack growth analysis for a crack growing laterally from the feed through hole show that

during 179,358 cycles, the rogue crack grows to a length of 0.148 inches. The inspection intervals based

on the center hole are therefore, not considered critical.

The following data is extracted from the AFGROW output file.

AFGROW 5.2.2.18 2/10/2015 17:26



Title: Longitudinal Loading, Lateral Cracks, Feedthru Hole

Load: Axial Stress Fraction: 1, Bending Stress Fraction: 0, Bearing Stress Fraction: 0 Advanced Models Thickness : 0.070 Width : 17.280




DOA EASA.21J.560



Page 55 of 109







Determine Stress State automatically (2 = Plane stress, 6 =

Plane strain) No K-Solution Filters


Plane strain fracture toughness: 29 Plane stress fracture toughness: 58 Effective fracture toughness for surface/elliptically shaped crack: 41 Fit parameters (KC versus Thickness Equation): Ak= 1, Bk=1 Yield stress: 48 Lower 'R' value boundary: -0.3 Upper 'R' value boundary: 0.7 Exponents in NASGRO Equation: n=2.601, p=0.5, q=1 Paris crack growth rate constant: 2.44e-008 Threshold stress intensity factor range at R = 0: 2.9 Threshold coefficient: 1.5 Plane stress/strain constraint factor: 1.5 Ratio of the maximum applied stress to the flow stress: 0.3

Residual stress: 4.931

Cycle by cycle beta and spectrum calculation


DOA EASA.21J.560



Page 56 of 109

Condition (i) Fuselage Loading Due to Bending and Pressure - Lateral Cracking

The critical areas for crack initiation and growth are between rivets in the outer row of rivets and cracks

emanating from the feed through hole.

The analysis considers a crack that grows from the edge of one rivet to the edge of the adjacent rivet in

the same row. These critical crack locations are illustrated below in Figure 5.2A. Only the skin is checked

as the skin has a higher stress than the doubler, the inspection of the doubler shall follow the same intervals

and techniques as the skin.

**Spectrum Information

Constant amplitude loading Spectrum multiplication factor: 4.277 SPL: 0

The spectrum will be repeated up to 10000000 times

Total Cycles: 1 Levels: 1 Subspectra: 1 Max Value: 1 Min Value: 0

No Spectrum Filters

Transition will be based on K max or 95% thickness penetration

Criteria

Length Beta Tension Beta Compression R(k)

R(final) Delta-K D( )/DN

Crack #1 Left Tip C 0.05 2.8506 2.8506

0.0000 0.0000

4.8321e+000 2.7616e-007

Right Tip C 0.01 3.2545 3.2545 0.0000 0.0000

2.4672e+000 0.0000e+000 Left Tip A 0.07 2.9560 2.9560 0.0000

0.0000 5.9287e+000 5.3944e-007

Right Tip A 0.07 0.6886 0.6886 0.0000 0.0000

1.3812e+000 0.0000e+000 Max stress 4.277, r = 0.00, 0 Cycles, Constant amp.: 1, Pass:

1

*********Transition at 95% thickness penetration



Crack #1 Left Tip C 0.05 2.9384 2.9384

0.0000 0.0000

4.9809e+000 3.0607e-007

Right Tip C 0.01 3.3822 3.3822 0.0000 0.0000

2.5640e+000 0.0000e+000 Max stress 4.277, r = 0.00, 0 Cycles, Constant amp.: 1, Pass:

1

. Length Beta Tension Beta Compression R(k)


Crack #1 Left Tip C 0.14 2.3422 2.3422

0.0000 0.0000

6.6435e+000 7.7010e-007

Right Tip C 0.01 3.4948 3.4948 0.0000 0.0000

2.6493e+000 0.0000e+000 Max stress 4.277, r = 0.00, 169166 Cycles, Constant amp.

: 169167,

Pass: 169167



Crack #1 Left Tip C 0.14798 2.3027 2.3027

0.0000 0.0000

6.7154e+000 7.9623e-007

Right Tip C 0.01 3.5046 3.5046 0.0000 0.0000

2.6568e+000 0.0000e+000 Max stress 4.277, r = 0.00, 179358 Cycles, Constant amp.

: 179359,

Pass: 179359

Stress State in 'C' direction (PSC): 6

Cycle count exceeded stop value - run time : 0 hour(s) 0 minute(s) 2 second(s)


DOA EASA.21J.560



Page 57 of 109

Joint Strength

The fatigue analysis of the reinforced skin is based on the Neutral Line Analysis (NLA), Reference Schijve,

et al [9,19], described in the appendix A of this report. This method is performed in two steps. The first is to calculate the fastener forces in the rivet rows used to attach the doubler to the skin. The second is to

calculate the secondary bending of the skin doubler system using the method presented by de Rijck, et al

[18]. Swift’s fastener flexibility model [20] is employed in the analysis. The geometry for this skin-doubler model is shown in Figure 5.2B.

The first two rivet rows on the LH side and the last two rivet rows on the RH side are modeled in the Neutral

Line Model. NAS1097AD4 rivets are installed in each of these rows with a rivet row spacing and rivet pitch as shown in the above Figure 5.2B.

Skin tension = (0.070-in)(9,295 psi*) = 650.65 lbs/in

*Ref: Section 3.1.2, Lateral LEFM stress for Fatigue

The analysis of the riveted joints uses the displacement-compatibility model shown in Figure 5.2C, with Swift’s fastener flexibility.


DOA EASA.21J.560



Page 58 of 109

The stiffnesses of the skin and doubler are calculated using the formula

where t represents the thickness, 0.070-in and 0.080 in for the skin and doubler, respectively; W representes the width of the douber, W = 17.0 in (Conservative Consideration); and the elementlength L

is the distance between rivet rows. This element length is also assigned to the “skin” elements to the left

between points 0 and SL3 and between points SR3 and N.

The compliance for Swift’s fastener model is defined for aluminum rivets as

where D represents the fastener diameter, t1 and t2 represent the skin and doubler thickesses, defined

above, and E = 10.5 (10)6 psi (Ref. Table 3.2.4.0(e1) MMPDS-06, represents Young’s modulus for the skin,

doubler and fastener. The resulting stiffness for a row of fasteners is calculated as

where Nf represents the number of fasteners in the row, thus Nf = 2 and Nf = 2 for the first and fourth forward rows of rivets repectively. The applied load P, is defined in terms of the longitudinal stress as

where the minus sign denotes the fact that the force acts in the –x direction. This load is used to calculate

displacements are calculated from the assembled stiffness matrix.

The maximun displacement occurs at point 0 and was calculated as u0 = 8.852381 x 10-4 in. The resulting

fastener loads are tabulated as


DOA EASA.21J.560



Page 59 of 109

The greatest bearing stress is calculated as

brg = 142.599 lb / [(0.070-in)(0.163 in*)] = 12,498 psi

*Ref: consider a hole size of 0.163-inch for the 1/8 rivet in a 0.129-in hole which has a head diameter of

0.196-in.

The corresponding bearing stress ratio kb is calculated as

kbrg = 12,498 psi / 9,295 psi = 1.345

The skin and doubler Stress resultants are reported in Tables 5.2B and 5.2C.

The secondary bending stresses generated by the doubler were analyzed using the Neutral Line Method suggested by Schijve [9,19]. This particular analysis incorporates the effects of fastener flexibility as

described by de Rijck [18].


DOA EASA.21J.560



Page 60 of 109

The corrective moments were computed as described in References 11 & 12 and are summarized in Table

5.2D.

The neutral line displacement solution is presented in Figure 5.2E, where the units of both axes are inches.

The bending stress is calculated from the differential moment-deflection equation

which can be written in terms of the homogeneous portion of the neutral line solution as

Substituting this moment into the bending stress equation for a flat plate

b = 6 (331.83 in-lb) / [(17.0 in)(0.070-in + 0.080-in)2] = 5,205 psi

The corresponding bending stress ratio kb is calculated as

kb = 5,205 psi / 9,295 psi = 0.560

The bypass stress is calculated as

brg = (491.27 lbs/in)*/0.070-in = 7,018 psi

*Ref: Table 5.2B


DOA EASA.21J.560



Page 61 of 109

The corresponding bypass stress ratio kbypass is calculated as

kbypass = 7,018 psi / 9,295 psi = 0.755

Fatigue Life

The fatigue life is considered using three methods and the most conservative result is utilized as the fatigue

based threshold.

Method 1

Method 1 is the 'Existing Life Limit' approach, which considers existing OEM life limitations placed on the

detail, assembly, or installation.

No life limits are listed in the applicable Airworthiness Limitations sections of the approved aircraft manual.

However, LOV of the 757-200 airplane is 50,000 FC per 14 CFR Paragraph 121.1115(f) amendment 25-

360. Therefore a conservative Design Service Goal (DSG) of 50,000 cycles is considered, in the fatigue

analysis.

DSG (1) = 50,000 cycles

Threshold Inspection (1) = 50,000 cycles/2 = 25,000 cycles

Recurring Inspection (1) = 25,000 cycles/2 = 12,500 cycles

Method 2

By applying the following series of managed scatter factors as shown in Scatter Reduction Factor* methods A and B that represents the alteration/installation on the aircraft, the fatigue life of the alteration

is determined. *Reference Patrick Safarian, DTA Seminar, Lesson 19, Spirit Aviation, Inc., Wichita, KS,

Feb 25-27, 2013. [Ref 15]







Reliability Factor* 0.48/Alum, 0.26/Steel*; A conversion factor to obtain a reliable life value from mean or characteristic life data.


Scale Factor 0.33 < F < 0.50*; A factor to adjust the design life value based on the percentage of details in the specimen to the number of detail in the actual structure.



DOA EASA.21J.560



Page 62 of 109


Note, the life of fatigue life of the subject is reduced by multiplying the MMPDS fatigue life by this Scatter

Reduction Factor.



Scale Factor 1.0 < F < 2.0*; To account for differences in scale and fidelity of the test, including the extent to which the loading of the test article represents the actual structure.


fidelity of the data.

Reliability Factor 2.75/alum, 3.5/Steel; A conversion factor to obtain a reliable life value from mean or

characteristic life data.



Note, the life of fatigue life of the subject is reduced by dividing the MMPDS fatigue life by this Scatter Reduction Factor.

Method 2 is the 'Fokker empirical prediction method' (Reference 9, Chapter 18). The stress concentration

factors for tension, bearing, and bending are calculated for the loaded rivet hole. Predictions are extrapolated from this curve by accounting of three contributions to the stress concentration at the rivet

holes of the critical end row. The contributions are associated with (i) load transmission by the rivet (pin

loading on hole), (ii) bypass loading of the rivet rows, and (iii) increased stress by secondary bending. The equation used is the following:

In this equation y is the percentage of the load transmitted to the other sheet in the critical row. Then, (1-

y) is the percentage of the bypass load. The factor k is the secondary bending factor.

Y = (Rivet Load/Rivet Pitch)/(Skin Tension)

= (142.599 lbs/0.90-in)/(650.65 lbs/in) = 0.244

As described in Peterson, Ref. 12. The case of a pinned joint in an infinite thin element has been solved

mathematically by Bickley (1928). The finite-width case has been solved by Knight (1935), where the element width is equal to twice the hole diameter d and by Theocaris (1956) for d/H = 0.2 to 0.5.

Experimental results (strain gage or photoelastic) have been obtained by Coker and Filon (1931),

Schaechterle (1934), Frocht and Hill (1940), Jessop, Snell and Holister (1958), and Cox and Brown (1964). (See Peterson for descriptions of these references).


DOA EASA.21J.560



Page 63 of 109

Nominal stress based on bearing area:

Nominal stress based on bearing area:

In Figure 5.2E, (Chart 4.67) the K t n b curve corresponds the Theocaris (1956) data for d/H = 0.2 to 0.5.

The values of Frocht and Hill (1940) and Cox and Brown (1964) are in good agreement with Chart 4.67,

although slightly lower. From d/H = 0.5 to 0.75 the foregoing 0.2-0.5 curve is extended to be consistent with the Frocht and Hill values. The resulting curve is for joints where c/H is 1.0 or greater. For c/H = 0.5,

the Ktn values are somewhat higher.

From Eq. (4.85), Ktnd = Ktnb at the d/H = 'A It would seem more logical to use the lower (full line) branches of the curves in Figure 5.2G (Chart 4.67), since, in practice; d/H is usually less than 'A This means that Eq.

(4.84), base on the bearing area, is generally used.

The rivets are considered countersunk in the doubler. This analysis will consider an initial through hole of

the average skin hole of 0.163 inches. The geometry concentration factor for the pin (Kt, pin) (i) is determined from the following graph, Peterson Chart 4.67:


DOA EASA.21J.560



Page 64 of 109

This data results in following “net” stress concentration factor.

Ktn, pin = 1.30

The geometry concentration factor for the hole in tension (Ktg, hole) (ii) is determined from Peterson Chart

4.1:


DOA EASA.21J.560



Page 65 of 109

This data results in the following “net” stress concentration factor.

Ktn= 2.56

The geometry concentration factory for the hole in bending (Ktg, bending) (iii) is determined from Peterson



DOA EASA.21J.560



Page 66 of 109

Ktn, hole bending = 1.68

The ratio of the bending stress to the applied tension stress (k) (iii) is obtained:

kb=0.560*

*Ref. below Figure 5.2D

The total stress concentration factor due to pin load, bypass, and secondary bending is denoted:

Kt= (γ)(Kt, pin) + (1- γ)(Kt, hole, tension)+ (kb)(Kt, hole bending)

Ktn= (0.244)(1.30) + (1–0.244)(2.56) + (0.560)(1.68) =3.19


DOA EASA.21J.560



Page 67 of 109

The Ktn value that is calculated is input into the S/N data curve presented in the MMPDS Ref. 1 to determine

the life of the installation. The life of the installation is calculated using a logarithmic interpolation.

Effective Stress in Kt= 2 panel:9,295psi

Seq= Smax (1- R)0.68 = 9.295ksi(1- 0)0.68 = 9.295ksi

log(Nf) = 9.2- 3.30[log(Seq– 8.5)] = 9.2- 3.30[log(9.295– 12.3)] =NONREAL

Nf= NEAR INFINITE≈ 1010


DOA EASA.21J.560



Page 68 of 109

Effective Stress in Kt = 4 panel: 9,295 psi

Seq = Smax(1-R)0.66 = 9.295 ksi(1-0)0.66 = 9.295 ksi

log(Nf) = 8.3 - 3.30[log(Seq – 8.5)] = 8.3 - 3.30[log(9.295 – 8.5)] = 8.628

Nf = 10(8.628)



DOA EASA.21J.560



Page 69 of 109

Fatigue Life = 10(Nf) = 10(9.1594) = 1.44 x 109 cycles


Fatigue life = (1.44 x 109 cycles) [(0.7a)(0.7b)(0.48c)(0.38d)] = 128,701,440 cycles aRef. Testing Factor bRef. Confidence Factor cRef. Reliability Factor1.5* dRef. Scale Factor


Fatigue life = (1.44 x 109 cycles)/[(1.5a)(1.5b)(2.75c)] = 232,727,272 cycles aRef. Scale Factor

bRef. Load Factor cRef. Reliability Factor

The recurrent inspection interval for Part 25 aircraft are not to be based on fatigue life per Patrick Safarian DTA Seminar, Feb 25-27, 2013. Fatigue life calculated with both A and B scatter factors are

acceptable. For this analysis the fatigue life with scatter factor of A is considered.

The acceptable conservative fatigue life limit for the longitudinal methods are therefore, based on Method 1.


Threshold Inspection = 50,000 cycles/2 = 25,000 cycles

Recurring Inspection = 25,000 cycles/2 = 12,500 cycles

Crack Growth Analysis for Crack between Rivets in Forward/Aft Row - Rogue Flaw


The simplified model for the lateral skin crack between two adjacent rivets in the side row of the skin is

modeled as a pin loaded hole in a flat sheet. The sheet width is taken to be the length of the skin bay. The

applied stress is the reference stress. The reference stress and applicable stress fractions are calculated at the beginning of this section.




DOA EASA.21J.560



Page 70 of 109

The residual strength curve is constructed by plotting the results of the AFGROW crack growth model for

the left 0.05 inch long initial crack. The 1000-9000 series aluminum, 2024-T3 Al, [Clad; plt & sht; T-L ] material properties from the AFGROW data base are used for the calculations.

E = 10,600 ksi υ = 0.33

Fty = 48 ksi,

KIC = 29 ksi√in

KC = 58 ksi√in

The applied fatigue remote stress and required residual strength are constant. The NSY curve is defined

by:

The Net Section Strength is determined by iteration (using a 0.001 convergence tolerance). The βC factor,

cycle count, and crack length for each crack growth increment are obtained from the AFGROW output data.


DOA EASA.21J.560



Page 71 of 109

The critical K value (apparent fracture toughness) is defined by:

Where:

index ~ Stress State Index (6 – plane strain, 2 – plane stress)

The “Allowable Stress” for KCrit is calculated as follows

The first intersection of the 'Residual Strength Required' line with either the KMAX or NSY lines indicates a

critical crack length and resulting failure of the detail.

The chart on the following page illustrates the residual strength characteristics of the model considering the Net Section Yield (NSY) and KMAX failure criterion.

Evaluation of the residual strength chart indicates that the right 0.05 inch initial crack has reached a critical

crack length of 0.71 inches base on Kmax criteria as it transitions to the adjacent rivet hole. Note that the


DOA EASA.21J.560



Page 72 of 109

non-linear nature of the NSY curve is due to the addition of the yield zone size to the crack length at the

crack tips.

The AFGROW crack growth curve for the 0.05 inch initial crack is shown on the next page. Note that the chart shown is for a crack growth model which propagates the crack to the next free edge, in the advanced

model case when a crack reaches an adjacent hole.

The AFGROW program also has options to halt the crack growth at a user specified crack length which is

used to determine the number of cycles required to reach a detectable crack length, and option to halt the program for NSY and KMAX failure criteria.

The first run of the program halts the crack growth at the user specified detectable crack length (with the

option selected to halt at a critical crack length). The second run of the program, using the same model, halts the program when either the NSY or KMAX criteria have been exceeded. A third run of the program is

conducted which allows the cracks to reach a free edge for illustrative purposes.


DOA EASA.21J.560



Page 73 of 109

The results of the crack growth analysis for a crack growing from one rivet hole to the adjacent rivet in the outer

row are shown on the next two pages. The analysis predicts 58,044 cycles are required to grow the crack to an inspectable length, (0.196-in – 0.129-in)/2 + 0.10 in = 0.1335-in.

The inspections are done from the outside of the airplane and has direct access to the skin. Therefore, the HFEC

(High Frequency Eddy Current) method can be used.


DOA EASA.21J.560



Page 74 of 109

The analysis predicts that 169,088 cycles are required to grow the crack to a length of 0.71 inches which is the

critical crack length.


DOA EASA.21J.560



Page 75 of 109

Threshold Inspection (1) = 58,044 cycles

Threshold Inspection (2) = 169,088 cycles / 2 = 84,544 cycles

Recurring Inspection (1) = (169,088 cycles - 58,044 cycles) / 3 = 37,014 cycles Recurring

Inspection (2) = 84,544 cycles / 2 = 42,272 cycles

The recommended threshold inspection and recurring inspection interval (for the lateral load case) are:

Threshold Inspection = 58,000 cycles

Recurring Inspection = 37,000 cycles


DOA EASA.21J.560



Page 76 of 109

The following data is extracted from the AFGROW output file:

AFGROW 5.2.2.18 2/10/2015 13:14



Title: Lateral Loading, Longitudinal Cracks, 1st Rivet Row

Load: Axial Stress Fraction: 0.755, Bending Stress Fraction: 0.56, Bearing Stress Fraction: 1.345 Advanced Models Thickness : 0.070 Width : 20.000




Hole #2 (Hole) Diameter = 0.163 Offset = 10





Determine Stress State automatically (2 = Plane stress,

6 = Plane strain) No K-Solution Filters


Plane strain fracture toughness: 29 Plane stress fracture toughness: 58 Effective fracture toughness for surface/elliptically shaped crack: 41 Fit parameters (KC versus Thickness Equation): Ak= 1, Bk=1 Yield stress: 48 Lower 'R' value boundary: -0.3


DOA EASA.21J.560



Page 77 of 109

Upper 'R' value boundary: 0.7 Exponents in NASGRO Equation: n=2.601, p=0.5, q=1 Paris crack growth rate constant: 2.44e-008 Threshold stress intensity factor range at R = 0: 2.9 Threshold coefficient: 1.5 Plane stress/strain constraint factor: 1.5 Ratio of the maximum applied stress to the flow stress: 0.3

Failure is based on the current load in the applied spectrum

Cycle by cycle beta and spectrum calculation


DOA EASA.21J.560



Page 78 of 109


DOA EASA.21J.560



Page 79 of 109

Crack Growth Analysis for Crack Extending from edge of feedthru - Rogue Flaw


The simplified model for the longitudinal skin crack between the feed through hole and the two adjacent

rivets in the skin is modeled as an open hole in a flat sheet. The sheet width is taken to be the length

of a skin bay, 20.0-in.

Applied stress to the skin at the feedthru hole = (415.9*lbs/in)/(0.070**in) = 5,941 psi *Ref

Table 5.2B **Ref: skin thickness

Residual Stress = (10,935*psi)x(5,941 psi)/(9,295**psi) = 6,989 psi

*Ref: Lateral residual stress, calculated in Section 3.2.1 **Ref: Lateral

LEFM stress, calculated in Section 3.2.1




DOA EASA.21J.560



Page 80 of 109

The results of the crack growth analysis for a crack growing longitudinally from the feed through hole show that

during 169,088 cycles, the rogue crack grows 0.63-in to the adjacent rivet holes. The inspection intervals based on the center hole are therefore, not considered critical.


DOA EASA.21J.560



Page 81 of 109

The following data is extracted from the AFGROW output file.


DOA EASA.21J.560



Page 82 of 109


DOA EASA.21J.560



Page 83 of 109

Fatigue Life - Stringers

The antenna installations drill holes in the existing stringer structures between FS 1440 and FS 1460.

The holes are drilled for the installation of MS20470AD6 rivets in the webs of the stringers.

The crack growth from the edge of a one fastener hole to the existing free edge of the stringer is

analyzed due to the tension load from the cabin pressurization and fuselage bending. These critical

crack locations are illustrated below in Figure 5.3A. The stress field is produced on the stringer due to

the cabin pressurization.

The loading of the stringers due to the cabin pressurization is calculated as suggested by Flhgge on NACA TN

2612 (section 1.1, ref 4). The fuselage radius of 81.0-in and the skin thickness of 0.070-in are considered for

the calculations. The calculations are shown in Figure 5.3D and 5.3E. The critical load factor applicable to the installation is cabin pressurization.

Pressure Associated with Lateral Stresses:

Residual Strength Stress:

P= (1.1(9.45 psi)) + 0.5* = 10.9 psi

*Ref. External Aerodynamic Pressure Cyclic LEFM Stress:

P= (9.45 psi)

The stringer are calculated using 7075-T6 aluminum and the properties are calculated using AutoCAD. The stringer cross-section is scaled from the LB Aircraft drawing. The frame cross-section varies around the

diameter of the fuselage. Since the axial stresses in the stringers are a function of the frame area, the analysis

conservatively considers a frame area equal to 0.8-in2 .

Stringer Spacing ≈ 8.64 in

Frame Spacing = 20.0-in


DOA EASA.21J.560



Page 84 of 109


DOA EASA.21J.560



Page 85 of 109

The effective fatigue stress in the stringers is very small and is well below the fatigue threshold for 7075-T6 aluminum, as shown in Figures 5.3E.


DOA EASA.21J.560



Page 86 of 109

Effective Stress in Kt = 4 panel: 1,706 psi

Seq = Smax(1-Rf51 = 1.706 ksi(1-0)°.51 = 1.706 ksi

log(Nf) = 10.2 - 4.63[log(Seq - 5.3)] = 10.2 - 4.63[log(1.706 - 5.3)] = NONREAL Nf =

NEAR INFINITE

Thus, the fatigue life is considered using the 'Existing Life Limit' approach, which considers existing

OEM life limitations placed on the detail, assembly, or installation.

No life limits are listed in the applicable Airworthiness Limitations sections of the approved aircraft

manual. However, LOV of the 757-200 airplane is 50,000 FC per 14 CFR Paragraph 121.1115(f)

amendment 25-360. Therefore a conservative Design Service Goal (DSG) of 50,000 cycles is

considered, in the fatigue analysis.

DSG (1) = Fatigue Life = 50,000 cycles



The working stress is too small to initiate a crack growth model for the modified stringers. Therefore,


DOA EASA.21J.560



Page 87 of 109

the inspection intervals for the stringers are established considering the DSG of the 757-200 airplane.

Fatigue Life - -11 Hat Section

During the “Not to Exceed” ultimate flight condition the ultimate stress in the -11 hat section is

34,527*psi, which occurs at the first row of rivets due to bending (*Ref below Figure 4.3D). The

working (fatigue) stress at this location during normal operating conditions is considered half of this

ultimate stress.

Fatigue Stress = (34,527 psi)/2 = 17,264 psi

The fatigue life is considered using two methods and the most conservative result is utilized as the

fatigue based threshold.

Method 1

Method 1 is the 'Existing Life Limit' approach, which considers existing OEM life limitations placed on

the detail, assembly, or installation.

No life limits are listed in the applicable Airworthiness Limitations sections of the approved aircraft

manual. However, LOV of the 757-200 airplane is 50,000 FC per 14 CFR Paragraph 121.1115(f)

amendment 25-360. Therefore the Design Service Goal (DSG) of 50,000 cycles is considered, in the

fatigue analysis.

DSG (1) = Fatigue Life = 50,000 cycles



Method 2

By applying the following series of managed scatter factors as shown in Scatter Reduction Factor*

methods A and B that represents the alteration/installation on the aircraft, the fatigue life of the

alteration is determined. *Reference Patrick Safarian, DTA Seminar, Lesson 19, Spirit Aviation, Inc.,

Wichita, KS, Feb 25-27, 2013. [Ref 15]







Reliability Factor* 0.48/Alum, 0.26/Steel*; A conversion factor to obtain a reliable life value from

mean or characteristic life data.



DOA EASA.21J.560



Page 88 of 109

Scale Factor 0.33 < F < 0.50*; A factor to adjust the design life value based on the percentage of

details in the specimen to the number of detail in the actual structure.



Note, the life of fatigue life of the subject is reduced by multiplying the MMPDS fatigue life by this

Scatter Reduction Factor.



Scale Factor 1.0 < F < 2.0*; To account for differences in scale and fidelity of the test, including the



fidelity of the data

Reliability Factor 2.75/alum, 3.5/Steel; A conversion factor to obtain a reliable life value from mean

or characteristic life data.



Note, the life of fatigue life of the subject is reduced by dividing the MMPDS fatigue life by this Scatter

Reduction Factor.

Method 2 is the 'Fokker empirical prediction method' (Reference 9, Chapter 18). The stress

concentration factors for tension, bearing, and bending are calculated for the loaded rivet hole.

Predictions are extrapolated from this curve by accounting of three contributions to the stress

concentration at the rivet holes of the critical end row. The contributions are associated with (i) load

transmission by the rivet (pin loading on hole), (ii) bypass loading of the rivet rows, and (iii) increased

stress by secondary bending. The equation used is the following: Kt _ (yXKt, pin) + (1 - Y)(Kt, hole, tension)+ (k)(Kt, hole bending)

In this equation y is the percentage of the load transmitted to the other sheet in the critical row. Then,

(1-y) is the percentage of the bypass load. The factor k is the secondary bending factor.

The MS20470D5 fastener hole is only considered to be loaded in bending therefore the above equation

for Kt is simplified:

Kt = Kt, hole bending

The geometry concentration factor for the hole in bending (Ktg, bending) (iii) is determined from Peterson



DOA EASA.21J.560



Page 89 of 109

This data results in the following “net” stress concentration factor.

*Ref. 0.9-in is the distance between rivets

Kt = Ktn, hole bending = 1.79


DOA EASA.21J.560



Page 90 of 109

Effective Stress in Kt = 1.5 panel: 17,264 psi

Stress in Kt = 1.5 panel: 17,264 psi

Seq = SmaxO-Rf66 = 17.264 ksi(1-0)0.66 = 17.264 ksi

log(Nf) = 7.5 - 2.13 [log(Seq)] = 7.5 - 2.13[log(17.264 - 23.7)] = NONREAL

Nf = NEAR INFINITE ~ 1010


DOA EASA.21J.560



Page 91 of 109

Effective Stress in Kt= 2 panel: 17,264psi

Seq= Smax(1- R)0.68 = 17.264ksi(1- 0)0.68 = 17.264ksi

log(Nf) = 9.2- 3.30[log(Seq– 8.5)] = 9.2- 3.30[log(17.264– 12.3)] =6.903

Nf= 10(6.903)



DOA EASA.21J.560



Page 92 of 109

Fatigue Life = 10(Nf) = 10(8-0657) = 116,3 3 2,2 1 5 cycles


Fatigue life = (1.16 x 108 cycles) [(0.7a)(0.7b)(0.48c)(0.38d)] = 1.03 x 107 cycles aRef. Testing

Factor bRef. Confidence Factor cRef. Reliability Factor dRef. Scale Factor


Fatigue life = (1.16 x 108 cycles)/[(1.5a)(1.5b)(2.75c)] = 1.87 x 107 cycles aRef. Scale Factor

bRef. Load Factor cRef. Reliability Factor

The recurrent inspection interval for Part 23 aircraft are not to be based on fatigue life per Patrick Safarian DTA

Seminar, Feb 25-27, 2013. Fatigue life calculated with both A and B scatter factors are acceptable. For this

analysis the fatigue life with scatter factor of A is considered.

The acceptable conservative fatigue life limit for the two methods are therefore, based on Method 1.


Threshold Inspection = 50,000 cycles/2 = 25,000 cycles Recurring

Inspection = 25,000 cycles/2 = 12,500 cycles


DOA EASA.21J.560



Page 93 of 109

Conclusion

The information contained herein supplements the basic Supplemental Structural Inspection Document only in those areas listed herein. For limitations and procedures consult the Instructions for Continued

Airworthiness, supplement to or the basic Airplane Maintenance Manuals.

Appendix A – Longitudinal Fatigue Loading (MathCAD Worksheets for Neutral Line Analysis)


DOA EASA.21J.560



Page 94 of 109


DOA EASA.21J.560



Page 95 of 109


DOA EASA.21J.560



Page 96 of 109


DOA EASA.21J.560



Page 97 of 109


DOA EASA.21J.560



Page 98 of 109


DOA EASA.21J.560



Page 99 of 109


DOA EASA.21J.560



Page 100 of 109


DOA EASA.21J.560



Page 101 of 109


DOA EASA.21J.560



Page 102 of 109


DOA EASA.21J.560



Page 103 of 109


DOA EASA.21J.560



Page 104 of 109


DOA EASA.21J.560



Page 105 of 109


DOA EASA.21J.560



Page 106 of 109


DOA EASA.21J.560



Page 107 of 109


DOA EASA.21J.560



Page 108 of 109


DOA EASA.21J.560



Page 109 of 109