Question

121 A circular aircraft window made from polycarbonate is clamped at its edge and will experience a uniformly distributed load due to the pressure difference between the cabin and the external environment. Specifications for the window are given in Table Q2 below. (b) If the pressure difference during flight is P 75 + 1 kPa, use the equation below to calculate the deflection, w, at the centre of the window and the uncertainty in this result 114] Deflection of circular membrane as a function of radius, r: 12 2 Pa4 Where, Eh3 12(1- v2) Window Data Young's Modulus Poisson's ratio Thickness Window Radius E-2611 GPa v 0.37t0.01 h-20.0+0.2 mm a- 30.0+0.1 cm DATA SHEET General formula fr error proagation is given by:δ9 -5x) +K +( .82

Answer 1

12(1 - v2)

26 100.02 12(1- 0.372)

D- 20082,65

at center r = 0

Pu 64D

75000 0.34 64 20082.65

Deflection :

4.7266 * 10-4 m w

Uncertainty:

first:

12(1 - v2)

121-J-7.724 10-7

3Eh2 3012397.173 Oh 12(1- v2)

EN *20 = 17218.352 12(1-1,2)2

av

$\delta E = 1GPa;\delta h = 0.2mm;\delta \nu = 0.01$

$(\Delta D) ^2= 596601.76+362981.47+29647.165$

$\Rightarrow \Delta D = 994.6$

now consider:

Pu 64D

$\frac{\partial w}{\partial P} = \frac{a^4}{64D} = 6.302*10^{-9}$

$\frac{\partial w}{\partial a} = \frac{4Pa^3}{64D} = 6.302*10^{-3}$

$\frac{\partial w}{\partial D} =- \frac{Pa^4}{64D^2} = -2.35*10^{-8}$

$(\Delta w) ^2=(\frac{\partial w}{\partial P}*\delta P)^2+(\frac{\partial w}{\partial a}*\delta a)^2+(\frac{\partial w}{\partial D}*\delta D)^2$

$(\Delta w) ^2= 3.97*10^{-11}+ 3.97*10^{-11}+5.479*10^{-10}$

$\Delta w = 2.505*10^{-5}$

4.7266 * 10-4 m w
and $\Delta w = 2.505*10^{-5}$