Question

x=10

A fabric used in air-inflated structures is subjected to a biaxial loading that results in normal stresses o x = (15*X) MPa and o, = (2*x+145) MPa. Knowing that the properties of the fabric can be approximated as E = 115 GPa and v = 0.25, determine the change in length of: A) Side AB B) Side BC C) diagonal AC y 100 mm 75 mm D 7777

Answer 1

Given:

$\sigma_x = 15 \times X = 15 \times 10 = 150 \; MPa \\ \sigma_z = 2X+145 = 2\times 10 + 145 = 165 \; MPa \\ E = 115 \; GPa\\ \nu = 0.25$

Part A.)

We know that,

$\delta_{AB} = l_{AB} \times \epsilon_x$

Where $\epsilon_x$ is Strain in x direction.

now,

$\epsilon_x = \frac{\sigma_x}{E} -\frac{\nu \sigma_z}{E} = \frac{150}{115}-\frac{0.25\times 165}{115} \\ \epsilon_x = 9.4565 \times 10^{-4}\\ \delta_{AB} = 100 \times 9.4565\times 10^{-4} = 0.09456\; mm$

PART B.)

We know that,

$\delta_{BC} = l_{BC} \times \epsilon_z$

Where $\epsilon_x$ is Strain in x direction.

now,

$\epsilon_z = \frac{\sigma_z}{E} -\frac{\nu \sigma_x}{E} = \frac{165}{115}-\frac{0.25\times 150}{115} \\ \epsilon_y = 11.0869 \times 10^{-4}\\ \delta_{BC} = 75 \times 11.0869\times 10^{-4} = 0.08315\; mm$

PART C.)

We will use Pythagoras theorem to fine change in length of diagonal AC:

$L_{AC} = \sqrt{L_{AB}^2+L_{BC}^2} = 125 mm$

$L_{AC}$ is initial length of diagonal before induction of stress.

Final length of diagonal after stress application is :

$L_{AC}^{'} = \sqrt{100.09456^2 + 75.08315^2} = 125.125538\;mm$

Hence, $\delta_{AC} = L_{AC}^{'}-L_{AC} = 0.125538\;mm$