Theoritical

Theoritical
Theoritical
Theoritical
Theoritical
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Substantiation for double shear joint

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Repair substantiation for R.OAEZ.000465-ER01

BOLT DATA Shear Bolt Material : Titanium Alloy : : : : : : : : : 0.75 * 0.58 * d / 4 =2

Reference 1

d A FoS u(bolt) y(bolt) a(bolt) u(bolt) y(bolt) a(bolt) I

= Shear Bolt Dia = Lug Hole Dia = Area along shear plane = Factor of Safety = Ultimate Tensile Strength = Yield Tensile Strength = Allowable Tensile Strength (Working Tensile strength) = Ultimate Shear Strength = Yield Shear Strength = Allowable Shear Strength (Working Shear strength) = Moment of Inertia of Bolt Circular Area

9.50 mm 70.88 mm2 1.50 896.32 Mpa 827.37 Mpa

u(bolt) u(bolt) y(bolt)

/ FoS =

597.55 Mpa 672.24 Mpa 479.88 Mpa 448.16 Mpa 399.82 mm4

( Reference 1) = ( Reference 1) =

u (bolt)

/ FoS =

* r4/4 =

LUG DATA Lug Material : Aluminum Alloy 4046 : : : : : (D - t 2 ) / 2 = = 4.87 mm 10.33 mm 18.65 mm 4.16 mm 37.50 mm 1.50 : : : : : : 0.75 * 0.58 * 240.00 Mpa 140.00 Mpa

t1 t2 D g w FoS u(lug) y(lug) a(lug) u(lug) y(lug) a(lug) Abr

= Thickness of the Lug =

Hold Open Rod Support End Fitting

= Distance between the lugs = Clearance between the Lugs = Width of the Lug = Factor of Safety = Ultimate Tensile Strength = Yield Tensile Strength = Allowable Tensile Strength (Working Tensile strength) = Ultimate Shear Strength = Yield Shear Strength = Allowable Shear Strength (Working Shear strength)

u(lug) u(lug) y(lug)

/ FoS =

160.00 Mpa 180.00 Mpa 81.20 Mpa 120.00 Mpa 46.27 mm2

( Reference 1) = ( Reference 1) =

u(lug)

/ FoS = d * t1 =

= Projected area where bearing pressure is applied

:

T/R LOAD DATA

W P

= Weight of the Thrust Reverser = Load on the Hold Open Rod Support Fitting

: : 9.81 * (W/3)=

791 Kgs 2586.57 N = 2.59 KN

MODES OF FAILURE 1) Failure by Shearing of Bolt 2) Failure by Bolt Bending 3) Failure by Tension or Compression through Bolt Hole 4) Failure by Shearing of the Lug 5) Failure by Bearing Stress

1)

FAILURE BY SHEARING OF BOLT Since the bolt shear loading condition is Double shear, it can be assumed that Load acting on each plane of shear is 1/2 the total load Maximum shear across the Shear Bolt Maximum shear across the Shear Bolt

Reference 2

= =

P ( Reference 2) 2A

=

18.246 N/mm2 18.25 Mpa

Since < a(bolt) (18.25 Mpa < 448.16 Mpa) The Design is considered to be safe

2)

FAILURE BY BOLT BENDING

Reference 3Analysis and Design of Flight Vehicle Structure - EF Bruhn Chapter D1.10

Bending moment distance

b

=

t1/ 2 + t2/ 4 + g ( Reference 3) =

9.1775 mm 11.87 N-m

Bending moment "M" = Bending Stress "b" =

M

*

r /I

=

56.38 N/mm2 56.38 Mpa

It is a general assumption that the allowable bending stress is 1/2 times the allowable normal stress Allowable Bending Stress "b(a)(bolt)" = 298.77 N/mm2 298.77 Mpa Since b P, the actual load (21.82 KN > 2.59 KN) The Design is considered to be safe

4)

FAILURE BY SHEARING OF LUG

Reference 5Analysis and Design of Flight Vehicle Structure - EF Bruhn Chapter D1.10

Ultimate Shear Strength u = P/A Hence the maximum design shear load, the lug can take is

P max = a(lug) * A Pmax =

a(lug) * (w-d) * t * 2 ==

32726.40 N 32.73 KN

The maximum load the bolt can take in tension or compression is Since the maximum shear load the Lug could withstand P max

> P, the actual load (32.73 KN > 2.59 KN) The Design is considered to be safe

5)

FAILURE BY BEARING STRESS

Reference 6Analysis and Design of Flight Vehicle Structure - EF Bruhn Chapter D1.10

Bearing Stress br =

P/Abr = =

27.95 N/mm2 27.95 Mpa

Since br