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AA-SM-010 Stress Due To Interference Fit Bushing Installation - Rev B

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Subject to restrictions on the cover or first page
Author: R. Abbott Document Number: AA-SM-010
Check: Revision Level : B
Date: 20/10/2013 Page: 1 of 2
Title: STANDARD SPREADSHEET METHOD
STRESS DUE TO INTERFERENCE FIT BUSHING INSTALLATION
(Abbott, Richard. Analysis and Design of Composite and Metallic Flight Vehicle Structures 1st Edition, 2016)
(AFFDL-TR-69-42, 1986)
The method to calculate built-in stresses caused by 'press fit' bushings will be taken from the US airforce
stress manual chapter 9. This gives the following strength criteria:
For Stress corrosion of the lug - the maximum tensile stress should not exceed 50% of Fty

A = Inner Radius of Bushing


B = Outer Radius of Bushing
C = Outer Radius of Ring
D = Inner Radius of Ring

Ring (Lug): Bushing: Max Interference:


D= 0.9975 in B= 1.00025 in I= 0.00275 in
C= 1.5 in A= 0.5 in
E = 10700000 psi E = 28500000 psi
μ= 0.33 μ= 0.3
Fty = 56000 psi Ftu = 129000 psi
Fty/2 = 28000 psi

Internal Pressure (p):


p = I / (D / E × ((C² + D²) / (C² - D²) + μ) + - B / E × ((A² + B²) / (A² - B²) + μ))
= 0.00275 / (0.998 / (1.07E+07) × ((1.5² + 0.998²) / (1.5² - 0.998²) + 0.33) + - 1 /
(2.85E+07) × ((0.5² + 1²) / (0.5² - 1²) + 0.3))
p= 8600.2 psi

Lug Radial Stress: Lug tangential stress :


Fr = - p Ft = p × ((C² + D²) / (C² - D²))
= -8600.2 psi = 8600 × ((1.5² + 0.998²) / (1.5² - 0.998²))
= 22237.2 psi
Lug shear Stress:
Fs = ((-8600) - 22237) / 2
= -15418.7 psi

MS (lug ) = 28000 / 22237 - 1 = 0.26


Bush tangential Compression Stress:
Ft = 2 × p × B² / (B² - A²)
= 2 × 8600 × 1² / (1² - 0.5²)
= 22929.9 psi
MS (bush) = 129000 / 22930 - 1 = 4.63

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Subject to restrictions on the cover or first page

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